JEE Permutations & Combinations Questions

Every 2-digit number having greatest common divisor $$6$$ with another 2-digit number must itself be divisible by $$6$$, because any common divisor of the two numbers divides their gcd.
Hence write the numbers as $$m = 6a$$ and $$n = 6b$$ with $$a,b \in \mathbb{N}$$.

Because $$m$$ and $$n$$ are 2-digit numbers,

$$10 \le m = 6a \le 99 \quad \text{and} \quad 10 \le n = 6b \le 99$$

which gives

$$\frac{10}{6} \le a \le \frac{99}{6}, \qquad \frac{10}{6} \le b \le \frac{99}{6}$$

or, after taking integer values,

$$2 \le a \le 16, \qquad 2 \le b \le 16.$$

The condition $$\gcd(m,n)=6$$ becomes $$\gcd(6a,6b)=6,$$ i.e.

$$\gcd(a,b)=1.$$ Thus we must count ordered pairs $$(a,b)$$ such that

$$2 \le a \lt b \le 16 \quad \text{and} \quad \gcd(a,b)=1.$$ Call such pairs “coprime pairs”.

List each value of $$a$$ and count suitable $$b$$ values individually:

$$\begin{aligned} a=2 &: b=3,5,7,9,11,13,15 &\Rightarrow 7\\ a=3 &: b=4,5,7,8,10,11,13,14,16 &\Rightarrow 9\\ a=4 &: b=5,7,9,11,13,15 &\Rightarrow 6\\ a=5 &: b=6,7,8,9,11,12,13,14,16 &\Rightarrow 9\\ a=6 &: b=7,11,13 &\Rightarrow 3\\ a=7 &: b=8,9,10,11,12,13,15,16 &\Rightarrow 8\\ a=8 &: b=9,11,13,15 &\Rightarrow 4\\ a=9 &: b=10,11,13,14,16 &\Rightarrow 5\\ a=10&: b=11,13 &\Rightarrow 2\\ a=11&: b=12,13,14,15,16 &\Rightarrow 5\\ a=12&: b=13 &\Rightarrow 1\\ a=13&: b=14,15,16 &\Rightarrow 3\\ a=14&: b=15 &\Rightarrow 1\\ a=15&: b=16 &\Rightarrow 1 \end{aligned}$$

Adding these counts:

$$7+9+6+9+3+8+4+5+2+5+1+3+1+1 = 64.$$

Therefore there are $$64$$ ordered pairs $$(a,b)$$, and hence $$64$$ ordered pairs $$(m,n)$$ with $$m \lt n$$ and $$\gcd(m,n)=6$$ where both $$m$$ and $$n$$ are 2-digit numbers.

Final answer: $$64$$.

Let $$a_n$$ denote the number of subsets of $$\{1,2,\ldots ,n\}$$ in which no two elements are consecutive.
We have to find $$a_5$$.

Step 1 Establish a recurrence for $$a_n$$.
Consider whether the element $$n$$ is present in a subset.

• Subsets that do not contain $$n$$: these are exactly the admissible subsets of $$\{1,2,\ldots ,n-1\}$$. The count is $$a_{\,n-1}$$.

• Subsets that do contain $$n$$: then $$n-1$$ cannot be present (to avoid consecutiveness). After fixing $$n$$, the remaining admissible part comes from $$\{1,2,\ldots ,n-2\}$$. The count is $$a_{\,n-2}$$.

Add the two mutually exclusive cases:
$$a_n = a_{\,n-1} + a_{\,n-2}$$ $$-(1)$$

Step 2 Set the initial values.
For $$n = 1$$: admissible subsets are $$\emptyset$$ and $$\{1\}$$, so $$a_1 = 2$$.
For $$n = 2$$: admissible subsets are $$\emptyset, \{1\}, \{2\}$$, so $$a_2 = 3$$.

Step 3 Use the recurrence to reach $$n = 5$$.

Using $$(1)$$:
$$a_3 = a_2 + a_1 = 3 + 2 = 5$$
$$a_4 = a_3 + a_2 = 5 + 3 = 8$$
$$a_5 = a_4 + a_3 = 8 + 5 = 13$$

Step 4 Conclusion.
Thus, the number of subsets of $$\{1,2,3,4,5\}$$ with no two consecutive elements is $$\boxed{13}$$.

First write the two series in compact sigma‐notation.

For $$a$$, every term after the first is of the form $$\frac{{}^{n}C_{2}}{(n+1)!}$$ with $$n \ge 2$$.
Hence $$a = 1 + \sum_{n = 2}^{\infty} \frac{{}^{n}C_{2}}{(n+1)!}\; -(1)$$

For $$b$$, the $$n^{\text{th}}$$ bracket gives $$\sum_{r = 0}^{n} {}^{n}C_{r}=2^{n}$$, so

$$b = 1 + \sum_{n = 1}^{\infty} \frac{2^{n}}{n!}\; -(2)$$

Case 1: Evaluation of $$a$$

Use $${}^{n}C_{2} = \frac{n(n-1)}{2}$$ in $$(1):$$

$$a = 1 + \sum_{n=2}^{\infty} \frac{n(n-1)}{2\,(n+1)!}$$

Simplify the general term:

$$\frac{n(n-1)}{(n+1)!} = \frac{n(n-1)}{(n+1)\,n\,(n-1)!} = \frac{1}{(n+1)(n-2)!}$$

Hence

$$a = 1 + \frac12 \sum_{n=2}^{\infty} \frac{1}{(n+1)(n-2)!}$$

Shift the index: put $$k = n-2 \;(k = 0,1,2,\dots)$$.

Then $$a = 1 + \frac12 \sum_{k=0}^{\infty} \frac{1}{(k+3)\,k!}\; -(3)$$

Use the identity $$\frac{1}{k+3} = \int_{0}^{1} x^{k+2}\,dx$$

Insert this into $$(3):$$

$$a = 1 + \frac12 \int_{0}^{1} \left[\sum_{k=0}^{\infty} \frac{x^{k}}{k!}\right] x^{2}\,dx = 1 + \frac12 \int_{0}^{1} x^{2} e^{x}\,dx$$

Integrate $$x^{2}e^{x}$$ once: $$\int x^{2} e^{x} dx = e^{x}(x^{2}-2x+2) + C$$

Therefore $$\int_{0}^{1} x^{2} e^{x} dx = e^{1}(1^{2}-2\cdot1+2) - e^{0}(0-0+2) = e(1) - 2 = e - 2$$

Put this back: $$a = 1 + \frac12 (e - 2) = \frac{e}{2}\; -(4)$$

Case 2: Evaluation of $$b$$

From $$(2)$$ and the Maclaurin series of $$e^{x}$$, $$b = 1 + \sum_{n=1}^{\infty} \frac{2^{n}}{n!} = \sum_{n=0}^{\infty} \frac{2^{n}}{n!} = e^{2}\; -(5)$$

Case 3: Required ratio

Using $$(4)$$ and $$(5):$$

$$\frac{2b}{a^{2}} = \frac{2\,e^{2}}{\left(\dfrac{e}{2}\right)^{2}} = \frac{2\,e^{2}}{e^{2}/4}=8$$

Hence $$\displaystyle \frac{2b}{a^{2}} = 8$$.

We have 20 lines: $$L_1, L_2, \ldots, L_{20}$$.

Odd-indexed lines $$L_1, L_3, L_5, \ldots, L_{19}$$ (10 lines) are all parallel to each other.

Even-indexed lines $$L_2, L_4, L_6, \ldots, L_{20}$$ (10 lines) all pass through a point P.

For maximum intersections, we count using $$\binom{20}{2}$$ total pairs and subtract cases where pairs don't intersect.

Total pairs: $$\binom{20}{2} = 190$$.

Pairs that don't give distinct intersection points:

1) Parallel lines (odd-indexed) don't intersect each other: $$\binom{10}{2} = 45$$ pairs with 0 intersection points.

2) Concurrent lines (even-indexed) all meet at point P: $$\binom{10}{2} = 45$$ pairs, but they all intersect at the same point P. So they contribute 1 point instead of 45 distinct points. We lose $$45 - 1 = 44$$ points.

Maximum intersection points = $$190 - 45 - 44 = 101$$.

The answer is $$\boxed{101}$$.

Total 3-digit numbers (no repetition): $$^5P_3=5\times4\times3=60$$

Now, for a number to be divisible by 3, the sum of its digits must be divisible by 3.

So, classifying digits by mod 3:

Mod 0: {3}

Mod 1: {4, 7}

Mod 2: {2, 5}

For sum ≡ 0 (mod 3), possible valid combination:

(0, 1, 2)

Number of ways of choosing 1 from each group:   $$1\times2\times2=4$$ ways (digit sets)

Each set can be arranged in: $$3!=6$$ ways

So total numbers divisible by 3:$$4\times6=24$$

We need to count 3-digit integers (100-999) whose digit sum equals 14.

Let the three-digit number have digits $$a$$ (hundreds), $$b$$ (tens), $$c$$ (units) where $$1 \leq a \leq 9$$, $$0 \leq b \leq 9$$, $$0 \leq c \leq 9$$, and $$a + b + c = 14$$.

Since $$a \geq 1$$, set $$a' = a - 1$$ so that $$0 \leq a' \leq 8$$ and the equation becomes $$a' + b + c = 13$$.

Without considering the upper bounds on $$a'$$, $$b$$, and $$c$$, the number of nonnegative integer solutions is given by stars and bars as $$\binom{13 + 2}{2} = \binom{15}{2} = 105$$.

To account for the constraints, subtract cases where one of the variables exceeds its bound using inclusion-exclusion.

If $$a' \geq 9$$, let $$a'' = a' - 9$$, giving $$a'' + b + c = 4$$ and $$\binom{4 + 2}{2} = 15$$ solutions. If $$b \geq 10$$, let $$b' = b - 10$$, giving $$a' + b' + c = 3$$ and $$\binom{3 + 2}{2} = 10$$ solutions. If $$c \geq 10$$, let $$c' = c - 10$$, giving $$a' + b + c' = 3$$ and $$\binom{3 + 2}{2} = 10$$ solutions. Any intersection of two or more such violations is impossible because it would require a sum exceeding 13.

Hence the total count is $$\text{Count} = 105 - 15 - 10 - 10 = 70$$.

The answer is 70.

We need to find the number of ways of getting a sum of 16 when throwing a die four times.

We need the number of solutions to $$x_1 + x_2 + x_3 + x_4 = 16$$ where each $$x_i \in \{1, 2, 3, 4, 5, 6\}$$.

Let $$y_i = x_i - 1$$, so $$0 \leq y_i \leq 5$$. The equation becomes:

$$ y_1 + y_2 + y_3 + y_4 = 16 - 4 = 12 $$

Without upper bound constraint: The number of non-negative integer solutions to $$y_1 + y_2 + y_3 + y_4 = 12$$ is:

$$ \binom{12 + 3}{3} = \binom{15}{3} = \frac{15 \times 14 \times 13}{6} = 455 $$

Subtract: at least one $$y_i \geq 6$$. If $$y_1 \geq 6$$, let $$z_1 = y_1 - 6 \geq 0$$. Then $$z_1 + y_2 + y_3 + y_4 = 6$$, giving $$\binom{9}{3} = 84$$ solutions. By symmetry, there are 4 such cases:

$$ \binom{4}{1} \times \binom{9}{3} = 4 \times 84 = 336 $$

Add back: at least two $$y_i \geq 6$$. If $$y_1 \geq 6$$ and $$y_2 \geq 6$$, let $$z_1 = y_1 - 6, z_2 = y_2 - 6$$. Then $$z_1 + z_2 + y_3 + y_4 = 0$$, giving $$\binom{3}{3} = 1$$ solution. There are $$\binom{4}{2} = 6$$ such pairs:

$$ \binom{4}{2} \times \binom{3}{3} = 6 \times 1 = 6 $$

Three or more $$y_i \geq 6$$: This would require the sum to be at least 18, which exceeds 12. So this is impossible.

$$ N = 455 - 336 + 6 = 125 $$

The answer is $$\boxed{125}$$.

Select 4 from Group A (4M, 5W) and 4 from Group B (5M, 4W) such that the total is 4 men and 4 women.

Let $$k$$ men be selected from Group A, which means $$4-k$$ women are chosen from Group A. Since the overall requirement is 4 men and 4 women, from Group B one must select $$4-k$$ men and $$k$$ women.

The parameter $$k$$ must satisfy $$0 \le k \le 4$$ for men in Group A, $$4-k \le 5$$ for women in Group A, $$4-k \le 5$$ for men in Group B, and $$k \le 4$$ for women in Group B. These conditions hold for $$k=0,1,2,3,4$$.

Therefore the total number of ways is given by

$$N = \sum_{k=0}^{4} \binom{4}{k}\binom{5}{4-k}\binom{5}{4-k}\binom{4}{k}$$

For $$k=0$$, the count is $$\binom{4}{0}\binom{5}{4}\binom{5}{4}\binom{4}{0} = 1 \times 5 \times 5 \times 1 = 25$$

For $$k=1$$, the count is $$\binom{4}{1}\binom{5}{3}\binom{5}{3}\binom{4}{1} = 4 \times 10 \times 10 \times 4 = 1600$$

For $$k=2$$, the count is $$\binom{4}{2}\binom{5}{2}\binom{5}{2}\binom{4}{2} = 6 \times 10 \times 10 \times 6 = 3600$$

For $$k=3$$, the count is $$\binom{4}{3}\binom{5}{1}\binom{5}{1}\binom{4}{3} = 4 \times 5 \times 5 \times 4 = 400$$

For $$k=4$$, the count is $$\binom{4}{4}\binom{5}{0}\binom{5}{0}\binom{4}{4} = 1 \times 1 \times 1 \times 1 = 1$$

Summing these gives $$N = 25 + 1600 + 3600 + 400 + 1 = 5626$$

The correct answer is 5626.

Alphabetical order of letters: E, G, N, T, T, W, Y. (Note: T is repeated twice).

Words starting with E: $$\frac{6!}{2!} = \frac{720}{2} = 360$$

Words starting with G: We now look for the word GTWENTY.

o Next letter is E: $$G E \dots$$ remaining (N, T, T, W, Y) $$\rightarrow \frac{5!}{2!} = 60$$

o Next letter is N: $$G N \dots$$ remaining (E, T, T, W, Y) $$\rightarrow \frac{5!}{2!} = 60$$

o Next letter is T: $$G T \dots$$

$$G T E \dots$$ remaining (N, T, W, Y) $$\rightarrow 4! = 24$$

$$G T N \dots$$ remaining (E, T, W, Y) $$\rightarrow 4! = 24$$

$$G T T \dots$$ remaining (E, N, W, Y) $$\rightarrow 4! = 24$$

$$G T W \dots$$ (This matches our word)

$$G T W E \dots$$

$$G T W E N \dots$$

$$G T W E N T Y$$ (This is the first word in this sequence) $$\rightarrow 1$$

Total Rank: $$360 + 60 + 60 + 24 + 24 + 24 + 1 = \mathbf{553}$$

Let $$a,\,b,\,c$$ denote the numbers of questions chosen from Sections A, B and C respectively.

Total questions to be attempted: $$a+b+c = 15$$.
Minimum requirement: $$a \ge 4,\; b \ge 4,\; c \ge 4$$.
Upper limits (because of the number of questions present): $$a \le 8,\; b \le 6,\; c \le 6$$.

Introduce new variables to handle the “at least 4” condition:
$$a = a' + 4,\; b = b' + 4,\; c = c' + 4,\qquad a',b',c' \ge 0$$.

Substituting in $$a+b+c=15$$ gives
$$a' + b' + c' = 15 - 12 = 3 \qquad -(1)$$

Equation $$(1)$$ demands non-negative integer solutions of sum $$3$$. These are:

• $$\,(3,0,0)$$ and its permutations  • $$\,(2,1,0)$$ and its permutations  • $$\,(1,1,1)$$.

Now apply the upper bounds.

Case 1: $$(a',b',c') = (3,0,0)$$
Then $$(a,b,c)=(7,4,4)$$ which satisfies $$a\le 8,\;b\le 6,\;c\le 6$$.
    The other two permutations give $$(a,b,c)=(4,7,4)$$ or $$(4,4,7)$$, both violating $$b\le6$$ or $$c\le6$$.
    Hence only one admissible triple: $$(7,4,4)$$.

Case 2: $$(a',b',c') = (2,1,0)$$
    All six permutations obey the limits, giving
    $$(6,5,4),\;(6,4,5),\;(5,6,4),\;(4,6,5),\;(5,4,6),\;(4,5,6).$$

Case 3: $$(a',b',c') = (1,1,1)$$
    This single solution yields $$(5,5,5)$$, within the limits.

Thus the admissible $$(a,b,c)$$ triples are

$$(7,4,4),\;(6,5,4),\;(6,4,5),\;(5,6,4),\;(4,6,5),\;(5,4,6),\;(4,5,6),\;(5,5,5).$$

For each triple, multiply the individual combinations:
  Section A: choose $$a$$ out of $$8 \;\;(\binom{8}{a})$$
  Section B: choose $$b$$ out of $$6 \;\;(\binom{6}{b})$$
  Section C: choose $$c$$ out of $$6 \;\;(\binom{6}{c})$$.

Compute every case:

$$\begin{aligned} (7,4,4):&\;\binom{8}{7}\binom{6}{4}\binom{6}{4}=8\cdot15\cdot15=1800\\[2pt] (6,5,4):&\;\binom{8}{6}\binom{6}{5}\binom{6}{4}=28\cdot6\cdot15=2520\\[2pt] (6,4,5):&\;28\cdot15\cdot6=2520\\[2pt] (5,6,4):&\;\binom{8}{5}\binom{6}{6}\binom{6}{4}=56\cdot1\cdot15=840\\[2pt] (4,6,5):&\;\binom{8}{4}\cdot1\cdot6=70\cdot6=420\\[2pt] (5,4,6):&\;56\cdot15\cdot1=840\\[2pt] (4,5,6):&\;70\cdot6\cdot1=420\\[2pt] (5,5,5):&\;56\cdot6\cdot6=2016 \end{aligned}$$

Add the eight results:
$$1800+2520+2520+840+420+840+420+2016 = 11376.$$

Hence, a student can select the required 15 questions in $$\mathbf{11376}$$ different ways.

"DISTRIBUTION" has 12 letters: D(1), I(3), S(1), T(2), R(1), B(1), U(1), O(1), N(1) — 9 distinct letters.

We count 4-letter words by cases based on repetition.

Case 1: All different. Choose 4 from 9 distinct letters: $$\binom{9}{4} \times 4! = 126 \times 24 = 3024$$.

Case 2: One pair + 2 different. Pairs available: I or T (2 choices). Choose 2 more from remaining 8: $$2 \times \binom{8}{2} \times \frac{4!}{2!} = 2 \times 28 \times 12 = 672$$.

Case 3: Two pairs. Choose 2 from $$\{I,T\}$$: $$\binom{2}{2} = 1$$. Arrangements: $$\frac{4!}{2!2!} = 6$$.

Case 4: Three same. Only I has 3 copies. Choose 1 more from 8: $$8 \times \frac{4!}{3!} = 32$$.

Total = $$3024 + 672 + 6 + 32 = \boxed{3734}$$.

We need to evaluate $$\sum_{r=1}^{9}\frac{^{11}C_r}{r+1}$$ and express it as $$\frac{n}{m}$$ with $$\gcd(n,m) = 1$$, then find $$n + m$$.

Use the identity $$\frac{\binom{n}{r}}{r+1} = \frac{1}{n+1}\binom{n+1}{r+1}$$.

This identity comes from:

$$\frac{\binom{n}{r}}{r+1} = \frac{n!}{r!(n-r)!(r+1)} = \frac{1}{n+1} \cdot \frac{(n+1)!}{(r+1)!(n-r)!} = \frac{1}{n+1}\binom{n+1}{r+1}$$

With $$n = 11$$:

$$\frac{\binom{11}{r}}{r+1} = \frac{1}{12}\binom{12}{r+1}$$

Substitute and evaluate the sum.

$$\sum_{r=1}^{9}\frac{\binom{11}{r}}{r+1} = \frac{1}{12}\sum_{r=1}^{9}\binom{12}{r+1} = \frac{1}{12}\sum_{k=2}^{10}\binom{12}{k}$$

where $$k = r + 1$$.

Use the binomial sum.

We know: $$\sum_{k=0}^{12}\binom{12}{k} = 2^{12} = 4096$$

$$\sum_{k=2}^{10}\binom{12}{k} = 4096 - \binom{12}{0} - \binom{12}{1} - \binom{12}{11} - \binom{12}{12}$$

$$= 4096 - 1 - 12 - 12 - 1 = 4070$$

Final calculation.

$$\frac{n}{m} = \frac{4070}{12} = \frac{2035}{6}$$

Check: $$\gcd(2035, 6)$$. $$2035 = 5 \times 407 = 5 \times 11 \times 37$$. $$6 = 2 \times 3$$. No common factors, so $$\gcd = 1$$.

$$n + m = 2035 + 6 = 2041$$

The answer is 2041.

Using standard binomial identities:

$$\alpha = \sum_{r=0}^n (4r^2+2r+1)\binom{n}{r} = 4[n(n+1)2^{n-2}] + 2[n2^{n-1}] + 2^n = 2^n(n+1)^2$$

$$\beta = \sum_{r=0}^n \frac{\binom{n}{r}}{r+1} + \frac{1}{n+1} = \frac{2^{n+1}-1}{n+1} + \frac{1}{n+1} = \frac{2^{n+1}}{n+1}$$

Divide the two expressions:

$$\frac{2\alpha}{\beta} = \frac{2 \cdot 2^n(n+1)^2}{\frac{2^{n+1}}{n+1}} = (n+1)^3$$

Given $$140 < (n+1)^3 < 281$$, the only matching perfect cube is $$6^3 = 216$$, so $$n+1 = 6 \implies n = 5$$.

We have 40 students who passed at least one of three subjects M, P, and C, with |M| = 20, |P| = 25, |C| = 16, |M ∩ P| ≤ 11, |P ∩ C| ≤ 15, and |M ∩ C| ≤ 15. We wish to determine the maximum number of students who passed all three subjects.

Since every student passed at least one subject, the Inclusion-Exclusion Principle gives

$$|M ∪ P ∪ C| = |M| + |P| + |C| - |M ∩ P| - |P ∩ C| - |M ∩ C| + |M ∩ P ∩ C| = 40$$

Substituting the known totals,

$$20 + 25 + 16 - |M ∩ P| - |P ∩ C| - |M ∩ C| + |M ∩ P ∩ C| = 40$$

which simplifies to

$$61 - \bigl(|M ∩ P| + |P ∩ C| + |M ∩ C|\bigr) + |M ∩ P ∩ C| = 40\,. $$

Letting $$S = |M ∩ P| + |P ∩ C| + |M ∩ C|$$ and $$t = |M ∩ P ∩ C|$$, this equation becomes

$$61 - S + t = 40\quad\Longrightarrow\quad S - t = 21\,,\quad\text{so}\quad S = 21 + t\,. $$

Each pairwise intersection is at most 11, 15, and 15 respectively, and must be at least $$t$$ since the triple intersection is contained in each pair. Hence the maximum possible sum is $$S=11+15+15=41$$, giving $$t ≤ 41 - 21 = 20$$, while also $$t ≤ |M ∩ P| ≤ 11$$. Thus $$t ≤ 11$$.

We must also ensure that every region in the Venn diagram has a non-negative number of students. For $$t = 10$$ we have $$S = 31$$ and may choose $$|M ∩ P| = 11\,,\; |P ∩ C| = 10\,,\; |M ∩ C| = 10\,. $$ Then

Only-M = $$20 - 11 - 10 + 10 = 9 \geq 0$$ $$\checkmark$$; Only-P = $$25 - 11 - 10 + 10 = 14 \geq 0$$ $$\checkmark$$; Only-C = $$16 - 10 - 10 + 10 = 6 \geq 0$$ $$\checkmark$$; (M ∩ P) only = $$11 - 10 = 1 \geq 0$$ $$\checkmark$$; (P ∩ C) only = $$10 - 10 = 0 \geq 0$$ $$\checkmark$$; (M ∩ C) only = $$10 - 10 = 0 \geq 0$$ $$\checkmark$$; and the total is $$9 + 14 + 6 + 1 + 0 + 0 + 10 = 40$$ $$\checkmark$$.

For $$t = 11$$ we would have $$S = 32$$. One might take $$|M ∩ P| = 11$$, $$|P ∩ C| = 11$$, $$|M ∩ C| = 10$$, giving Only-M = $$20 - 11 - 10 + 11 = 10 \geq 0$$, Only-P = $$25 - 11 - 11 + 11 = 14 \geq 0$$, Only-C = $$16 - 11 - 10 + 11 = 6 \geq 0$$, but then (M ∩ C) only = $$10 - 11 = -1 < 0$$ $$\times$$. Any other choice with $$|M ∩ P|=11$$ requires the other two intersections to sum to 21 while each is at least 11, forcing a sum of at least 22, which is impossible. Hence $$t = 11$$ cannot be realized.

These considerations show that the maximum feasible value of $$|M ∩ P ∩ C|$$ is 10.

The answer is 10.

$$\text{Total elements in } R = 2 + 8 + 6 + 3 + 4 + 2 = 25$$

The relation $$R$$ contains exactly 25 elements.

For each $$a \in A = \{1, 2, ..., 7\}$$, $$f(a)$$ must be a subset of $$A$$ that contains $$a$$.

For each element $$a$$, the number of subsets of $$A$$ containing $$a$$ is $$2^6 = 64$$ (the other 6 elements can each be in or out).

Since the choices for each element are independent, the total number of functions is $$64^7 = (2^6)^7 = 2^{42}$$.

We need $$m^n = 2^{42}$$ with $$m$$ being the least natural number base. The smallest such $$m$$ is $$m = 2$$, giving $$n = 42$$.

$$m + n = 2 + 42 = \boxed{44}$$.

Find the number of elements in $$S = \{(x,y,z): x,y,z \in \mathbb{Z}, x+2y+3z = 42, x,y,z \geq 0\}$$. We seek nonnegative integer solutions to $$x + 2y + 3z = 42$$. For each integer $$z$$ with $$0 \le z \le 14$$ (since $$3z \le 42$$), we set $$R = 42 - 3z$$ and count the nonnegative solutions to $$x + 2y = R$$. Because $$y$$ may range from $$0$$ to $$\lfloor R/2 \rfloor$$, and for each $$y$$ the corresponding $$x = R - 2y$$ is nonnegative, there are $$\lfloor R/2 \rfloor + 1$$ solutions for each fixed $$z$$.

Specifically, when $$z=0$$, we have $$R=42$$ and count $$21+1=22$$; when $$z=1$$, $$R=39$$ giving $$19+1=20$$; when $$z=2$$, $$R=36$$ giving $$18+1=19$$; when $$z=3$$, $$R=33$$ giving $$16+1=17$$; when $$z=4$$, $$R=30$$ giving $$15+1=16$$; when $$z=5$$, $$R=27$$ giving $$13+1=14$$; when $$z=6$$, $$R=24$$ giving $$12+1=13$$; when $$z=7$$, $$R=21$$ giving $$10+1=11$$; when $$z=8$$, $$R=18$$ giving $$9+1=10$$; when $$z=9$$, $$R=15$$ giving $$7+1=8$$; when $$z=10$$, $$R=12$$ giving $$6+1=7$$; when $$z=11$$, $$R=9$$ giving $$4+1=5$$; when $$z=12$$, $$R=6$$ giving $$3+1=4$$; when $$z=13$$, $$R=3$$ giving $$1+1=2$$; and when $$z=14$$, $$R=0$$ giving $$0+1=1$$ solutions.

Summing these counts for $$z=0,1,\dots,14$$ yields $$22+20+19+17+16+14+13+11+10+8+7+5+4+2+1 = 169$$. Therefore, the number of elements in $$S$$ is 169.

We first list all ordered pairs $$(a,b)$$ in $$S\times S$$ that satisfy the restriction $$|a-b|\ge 2$$.

Total ordered pairs in $$S\times S$$ are $$6\times 6 = 36$$.

Forbidden pairs:
• Diagonal pairs with $$a=b$$  → 6 pairs.
• Pairs with $$|a-b|=1$$.
   When $$b=a+1$$: $$(1,2),(2,3),(3,4),(4,5),(5,6)$$ → 5 pairs.
   When $$b=a-1$$: $$(2,1),(3,2),(4,3),(5,4),(6,5)$$ → 5 pairs.
So forbidden pairs = $$6+5+5 = 16$$.

Allowed pairs (with $$|a-b|\ge 2$$) = $$36-16 = 20$$.

Every relation $$R\in X$$ is a subset of these 20 allowed pairs containing exactly 6 elements. Choosing any 6 of the 20 allowed pairs gives such a relation, and every such choice is valid.

Hence $$n(X)=\binom{20}{6}= {}^{m}C_6$$ which implies $$m=20$$.

Final Answer: 20

The set $$S = \{1,2,3,4,5,6\}$$ has six elements. Every relation under consideration must satisfy the two fixed conditions:

(i) it contains exactly $$6$$ ordered pairs, and
(ii) each ordered pair $$(a,b)$$ in it obeys $$|a-b|\ge 2$$.

We study the two subsets of relations, $$Y$$ and $$Z$$, separately.

By definition, each relation $$R\in Y$$ has a range that is a single element, say $$\{b\}$$. Hence the only possible pairs in such an $$R$$ are of the form $$(a,b)$$ with $$a\in S$$. Because $$R$$ must contain exactly six distinct ordered pairs, every first component $$a=1,2,3,4,5,6$$ must occur once:

$$R=\{(1,b),(2,b),(3,b),(4,b),(5,b),(6,b)\}.$$

For this relation to be admissible, each pair must satisfy $$|a-b|\ge 2$$. In particular, $$|b-b|=0\lt 2$$ fails for the pair $$(b,b)$$. Therefore no value of $$b$$ can make all six pairs valid, and hence

$$n(Y)=0.$$

A relation $$R\in Z$$ is a function from $$S$$ to $$S$$. Thus for every $$a\in S$$ there is exactly one ordered pair $$(a,f(a))$$ in $$R$$. The requirement $$|a-f(a)|\ge 2$$ must hold for each $$a$$.

We list the permissible images for every element of $$S$$:

For $$a=1$$: $$|1-b|\ge 2\Longrightarrow b=3,4,5,6$$  ⇒  4 choices.
For $$a=2$$: $$|2-b|\ge 2\Longrightarrow b=4,5,6$$  ⇒  3 choices.
For $$a=3$$: $$|3-b|\ge 2\Longrightarrow b=1,5,6$$  ⇒  3 choices.
For $$a=4$$: $$|4-b|\ge 2\Longrightarrow b=1,2,6$$  ⇒  3 choices.
For $$a=5$$: $$|5-b|\ge 2\Longrightarrow b=1,2,3$$  ⇒  3 choices.
For $$a=6$$: $$|6-b|\ge 2\Longrightarrow b=1,2,3,4$$  ⇒  4 choices.

Because the six choices are independent (the function need not be one-to-one), the total number of such functions equals the product of the individual counts:

$$n(Z)=4\cdot 3\cdot 3\cdot 3\cdot 3\cdot 4 =4^{2}\,3^{4}=16\cdot 81=1296.$$

$$n(Y)+n(Z)=0+1296=1296=k^{2}\; \Longrightarrow\; |k|=\sqrt{1296}=36.$$

Hence the required value is 36.

The multiset of available digits is $$\{0,1,2,2,2,4,4\}$$.
A member of $$X$$ is a 5-digit permutation of five of these digits, the left-most digit being non-zero.

Step 1 : Condition “multiple of 5’’
A number is a multiple of 5 only when its last digit is $$0$$ (the digit $$5$$ is not present).
Fix the units place as $$0$$. The remaining four positions must be filled with the digits of the multiset
$$\{1,2,2,2,4,4\}$$.

Step 2 : Count of all numbers ending in 0
Let $$a,b,c$$ be the numbers of $$1\,'\!s, 2\,'\!s, 4\,'\!s$$ used in the first four places.
They satisfy $$a+b+c=4,\; 0\le a\le1,\;0\le b\le3,\;0\le c\le2.$$ For every triple $$(a,b,c)$$ the number of distinct arrangements is $$\dfrac{4!}{a!\,b!\,c!}.$$ Listing all admissible triples:

$$ \begin{array}{ccccl} a & b & c & \text{Digits used} & \text{Arrangements} \\ \hline 0 & 3 & 1 & 2,2,2,4 & \dfrac{4!}{3!\,1!}=4\\ 0 & 2 & 2 & 2,2,4,4 & \dfrac{4!}{2!\,2!}=6\\ 1 & 3 & 0 & 1,2,2,2 & \dfrac{4!}{1!\,3!}=4\\ 1 & 2 & 1 & 1,2,2,4 & \dfrac{4!}{1!\,2!\,1!}=12\\ 1 & 1 & 2 & 1,2,4,4 & \dfrac{4!}{1!\,1!\,2!}=12 \end{array} $$

Total 4-digit arrangements (hence total numbers ending with 0):
$$T = 4+6+4+12+12 = 38.$$

Step 3 : Extra condition for “multiple of 20’’
A number ending in $$0$$ is divisible by $$20$$ precisely when the tens digit (4th position) is even.
After fixing the unit digit as $$0$$ there is no second zero left, so the tens digit can only be $$2$$ or $$4$$.
Thus, the only disallowed case is when the tens digit equals $$1$$.

Case - Tens digit = 1
Choose $$1$$ for the 4th position. The remaining multiset for the first three places is $$\{2,2,2,4,4\}$$.
All 3-digit permutations from this multiset are:

• 2,2,2  (1 way)
• 2,2,4  (3!/2! = 3 ways)
• 2,4,4  (3!/2! = 3 ways)

Hence numbers whose tens digit is $$1$$:
$$N_1 = 1+3+3 = 7.$$

Step 4 : Count with even tens digit
Numbers satisfying the tens-digit condition (i.e. multiples of 20) are therefore
$$N_{\text{even}} = T - N_1 = 38 - 7 = 31.$$

Step 5 : Required conditional probability
Let $$p$$ denote the probability that a randomly chosen element of $$X$$ is a multiple of $$20$$ given that it is a multiple of $$5$$. By definition $$p = \dfrac{N_{\text{even}}}{T} = \dfrac{31}{38}.$$

Step 6 : Final value
$$38p = 38 \times \dfrac{31}{38} = 31.$$

Hence the required value is 31.

We need to find 3-digit numbers divisible by either 3 or 4 but not divisible by 48.

Range: 100 to 999. First multiple: 102, last: 999.

$$\text{Count} = \frac{999 - 102}{3} + 1 = \frac{897}{3} + 1 = 299 + 1 = 300$$

First multiple: 100, last: 996.

$$\text{Count} = \frac{996 - 100}{4} + 1 = \frac{896}{4} + 1 = 224 + 1 = 225$$

First multiple: 108, last: 996.

$$\text{Count} = \frac{996 - 108}{12} + 1 = \frac{888}{12} + 1 = 74 + 1 = 75$$

$$|A \cup B| = 300 + 225 - 75 = 450$$

First multiple: 144, last: 960.

$$\text{Count} = \frac{960 - 144}{48} + 1 = \frac{816}{48} + 1 = 17 + 1 = 18$$

$$450 - 18 = 432$$

The answer is Option B: $$\mathbf{432}$$.

We need to count numbers strictly between 5000 and 10000 formed using the digits 1, 3, 5, 7, 9 without repetition.

The numbers must be 4-digit numbers from 5001 to 9999. Since all available digits are odd and nonzero, every such arrangement gives a valid number.

For the number to be between 5000 and 10000, the thousands digit must be 5, 6, 7, 8, or 9. From our available digits {1, 3, 5, 7, 9}, the valid first digits are 5, 7, and 9 — giving 3 choices.

After choosing the first digit, we have 4 remaining digits and 3 remaining positions.

Number of ways = $$4 \times 3 \times 2 = 24$$

Total = $$3 \times 24 = 72$$

The correct answer is Option D: 72.

The letters of MONDAY are: M, O, N, D, A, Y. In alphabetical order: A, D, M, N, O, Y.

To find the position of MONDAY, we count how many words come before it in dictionary order.

Words starting with A or D (before M): $$2 \times 5! = 2 \times 120 = 240$$

Words starting with M, then a letter before O (i.e., MA..., MD..., MN...): $$3 \times 4! = 3 \times 24 = 72$$

Words starting with MO, then a letter before N (i.e., MOA..., MOD...): $$2 \times 3! = 2 \times 6 = 12$$

Words starting with MON, then a letter before D (i.e., MONA...): $$1 \times 2! = 2$$

Words starting with MOND, then a letter before A: none, since A is the smallest remaining letter. So $$0$$.

MONDAY itself is the next word: $$+1$$

Total position = $$240 + 72 + 12 + 2 + 0 + 1 = 327$$

We need to find the serial number of the word THAMS when all permutations of the letters M, A, T, H, S are arranged in dictionary order.

The sorted order of letters is: A, H, M, S, T.

Words starting with letters before T:

Letters A, H, M, S come before T. Each gives $$4! = 24$$ words.

Count = $$4 \times 24 = 96$$

Words starting with T, then letters before H:

Remaining letters: {A, H, M, S}. Only A comes before H.

Count = $$1 \times 3! = 6$$

Words starting with TH, then letters before A:

Remaining letters: {A, M, S}. No letter comes before A.

Count = $$0$$

Words starting with THA, then letters before M:

Remaining letters: {M, S}. No letter comes before M.

Count = $$0$$

The word THAMS itself: Count = $$1$$

Serial number = $$96 + 6 + 0 + 0 + 1 = 103$$

The answer is Option A: $$103$$.

Given: The word MATHEMATICS has 11 letters: M, A, T, H, E, M, A, T, I, C, S.

The repeated letters are: M appears 2 times, A appears 2 times, T appears 2 times. The remaining letters H, E, I, C, S each appear once.

Total arrangements:

$$\text{Total} = \frac{11!}{2! \times 2! \times 2!} = \frac{39916800}{8} = 4989600$$

Arrangements where C and S are together:

Treat C and S as a single unit. This gives 10 items to arrange (with M appearing 2 times, A appearing 2 times, T appearing 2 times). The unit CS can be arranged internally in $$2! = 2$$ ways (CS or SC).

$$\text{Together} = \frac{10!}{2! \times 2! \times 2!} \times 2! = \frac{3628800}{8} \times 2 = 453600 \times 2 = 907200$$

Arrangements where C and S do NOT come together:

$$\text{Not together} = 4989600 - 907200 = 4082400$$

We are given that this equals $$(6!)k$$:

$$k = \frac{4082400}{6!} = \frac{4082400}{720} = 5670$$

The correct answer is Option B: 5670.

The word INDEPENDENCE has 12 letters, of which the vowels are I, E, E, E, E (5 vowels with E repeated 4 times) and the consonants are N, D, P, N, D, N, C (7 consonants with N repeated 3 times and D repeated 2 times).

To count the arrangements in which all the vowels appear together, treat the group of vowels as a single unit. Along with the 7 consonants, this gives 8 units to arrange. Accounting for the repeated letters, the number of ways to arrange these units is $$\frac{8!}{3! \times 2!} = \frac{40320}{6 \times 2} = 3360$$.

Within the vowel group, the 5 letters can be arranged in $$\frac{5!}{4!} = 5$$ ways, since E is repeated 4 times.

Multiplying these yields the total number of arrangements: $$3360 \times 5 = 16800$$.

The correct answer is 16800.

Five-digit numbers > 40000, divisible by 5, using digits {0, 1, 3, 5, 7, 9} without repetition.

For divisibility by 5, the last digit must be 0 or 5.

Case 1: Last digit = 0

First digit ≥ 4: choices are {5, 7, 9} = 3 options

Remaining 3 positions: 4 × 3 × 2 = 24

Subtotal = 3 × 24 = 72

Case 2: Last digit = 5

First digit ≥ 4 and ≠ 5: choices are {7, 9} = 2 options

Remaining 3 positions from 4 remaining digits: 4 × 3 × 2 = 24

Subtotal = 2 × 24 = 48

Total = 72 + 48 = 120

We need integers greater than 7000 using digits 3, 5, 6, 7, 8 without repetition.

These can be 4-digit numbers (> 7000) or 5-digit numbers.

5-digit numbers: All 5-digit numbers formed using 3, 5, 6, 7, 8 without repetition are greater than 7000.

Count = $$5! = 120$$

4-digit numbers greater than 7000:

The first digit must be 7 or 8 (to make the number ≥ 7000).

First digit = 7: Remaining 3 digits chosen from {3, 5, 6, 8} in $$4 \times 3 \times 2 = 24$$ ways. But we need numbers > 7000, and any 4-digit number starting with 7 is ≥ 7000. Since 7000 itself can't be formed (no 0), all such numbers are > 7000.

Count = $$P(4,3) = 24$$

First digit = 8: Remaining 3 digits chosen from {3, 5, 6, 7} in $$4 \times 3 \times 2 = 24$$ ways.

Count = $$P(4,3) = 24$$

Total 4-digit numbers = $$24 + 24 = 48$$

Total = 120 + 48 = 168

The correct answer is Option 2: 168.

We need to find the number of triplets $$(x, y, z)$$ of distinct non-negative integers satisfying $$x + y + z = 15$$.

Total solutions without distinctness constraint.

The number of non-negative integer solutions of $$x + y + z = 15$$ is:

$$\binom{15 + 2}{2} = \binom{17}{2} = 136$$

Subtract solutions where at least two variables are equal.

Let A = {x = y}, B = {y = z}, C = {x = z}.

$$|A|$$: If $$x = y$$, then $$2x + z = 15$$, so $$x$$ ranges from 0 to 7. That gives 8 solutions. Similarly $$|B| = |C| = 8$$.

$$|A \cap B|$$: If $$x = y = z$$, then $$3x = 15$$, so $$x = 5$$. That gives 1 solution. Similarly for all pairwise intersections and the triple intersection: $$|A \cap B| = |A \cap C| = |B \cap C| = |A \cap B \cap C| = 1$$.

By inclusion-exclusion:

$$|A \cup B \cup C| = 8 + 8 + 8 - 1 - 1 - 1 + 1 = 22$$

Distinct solutions.

$$136 - 22 = 114$$

The number of triplets is 114.

We need 3-digit numbers using {1, 3, 5, 8} with repetition, divisible by 3.

Digits mod 3: 1→1, 3→0, 5→2, 8→2. Residue counts: 0→1 choice, 1→1 choice, 2→2 choices.

Valid residue triplets summing to 0 mod 3:

(0,0,0): 1×1×1 = 1

(0,1,2): 6 arrangements × 1×1×2 = 12

(1,1,1): 1×1×1 = 1

(2,2,2): 2×2×2 = 8

Total = 1 + 12 + 1 + 8 = 22

The answer is $$\mathbf{22}$$.

We need to seat 5 girls and 7 boys at a round table such that no two girls sit together.

Seating the 7 boys around a circle can be done in $$(7-1)! = 6! = 720$$ ways. This arrangement creates 7 gaps between consecutive boys.

We select 5 of these gaps for the girls in $$\binom{7}{5} = 21$$ ways. Then the 5 girls can be arranged among themselves in $$5! = 120$$ ways.

Multiplying these numbers gives the total arrangements as $$720 \times 21 \times 120 = 1814400$$.

This equals $$126 \times (5!)^2 = 126 \times 14400 = 1814400$$. Noting that $$\binom{7}{5} \times 6! = 21 \times 720 = 15120 = 126 \times 120 = 126 \times 5!$$, the answer can be succinctly written as $$126(5!)^2$$.

$$\frac{^{2n+1}P_{n-1}}{^{2n-1}P_{n}} = \frac{11}{21}$$

$$\frac{\frac{(2n+1)!}{(2n+1-(n-1))!}}{\frac{(2n-1)!}{(2n-1-n)!}} = \frac{11}{21} \implies \frac{(2n+1)!}{(n+2)!} \cdot \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$$

$$\frac{(2n+1)(2n)(2n-1)!}{(n+2)(n+1)(n)(n-1)!} \cdot \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$$

$$\frac{2(2n+1)}{(n+2)(n+1)} = \frac{11}{21} \implies 42(2n+1) = 11(n^2+3n+2)$$

$$84n + 42 = 11n^2 + 33n + 22 \implies 11n^2 - 51n - 20 = 0$$

$$(11n + 4)(n - 5) = 0 \implies n = 5$$

$$n^2 + n + 15 = 5^2 + 5 + 15 = 45$$

Just use basic PnC formula:-

$$^nC_r=\frac{n!}{r!\ \left(n-r\right)!}$$

$$\frac{\frac{2n!}{\left(2n-3\right)!3!}}{\frac{n!}{\left(n-3\right)!3!}}=\frac{10}{1}$$

$$\frac{2n\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{\left(2n-3\right)!\ 3!}=\frac{10\ n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{1\ \left(n-3\right)!\ 3!}$$

$$2n\left(2n-1\right)\left(2n-2\right)=10n\left(n-1\right)\left(n-2\right)$$

Solve it and get the value of n.

You will get n=1,8

1 will be rejected because n>=3 is allowed only.

Putting the n = 8 in the given equation you will get 2:1 as answer.

To solve this, we use the property of Cartesian products and combinations.

1. Find the total elements in $$A \times B$$

The number of elements in the Cartesian product $$A \times B$$ is:

$$n(A \times B) = n(A) \times n(B) = 5 \times 2 = 10$$

2. Set up the Combination Sum

We need the number of subsets having "at least 3 and at most 6" elements. This is the sum of choosing 3, 4, 5, and 6 elements out of 10:

$$\text{Number of subsets} = ^{10}C_3 + ^{10}C_4 + ^{10}C_5 + ^{10}C_6$$

3. Calculate each term

• $$^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$
• $$^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$
• $$^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$$
• $$^{10}C_6 = ^{10}C_4 = 210$$

4. Final Result

$$\text{Sum} = 120 + 210 + 252 + 210 = \mathbf{792}$$

We need to count the number of ways to choose distinct integers $$x$$ and $$y$$ with $$1 \leq x \leq 25$$ and $$1 \leq y \leq 25$$ such that $$x + y$$ is divisible by 5.

In $$\{1, 2, \ldots, 25\}$$, each residue class has exactly 5 elements:

• Residue 0: {5, 10, 15, 20, 25}
• Residue 1: {1, 6, 11, 16, 21}
• Residue 2: {2, 7, 12, 17, 22}
• Residue 3: {3, 8, 13, 18, 23}
• Residue 4: {4, 9, 14, 19, 24}

The valid residue pairs are: $$(0,0), (1,4), (2,3), (3,2), (4,1)$$.

• $$(0,0)$$: Choose $$x$$ and $$y$$ from the same 5 elements, $$x \neq y$$: $$5 \times 4 = 20$$
• $$(1,4)$$: $$5 \times 5 = 25$$
• $$(2,3)$$: $$5 \times 5 = 25$$
• $$(3,2)$$: $$5 \times 5 = 25$$
• $$(4,1)$$: $$5 \times 5 = 25$$

Total ordered pairs:

$$20 + 25 + 25 + 25 + 25 = 120$$

The answer is $$120$$.

Find the number of integral solutions to $$x + y + z = 21$$ where $$x \geq 1$$, $$y \geq 3$$, $$z \geq 4$$.

Substitute to remove lower bounds.

Let $$a = x - 1 \geq 0$$, $$b = y - 3 \geq 0$$, $$c = z - 4 \geq 0$$.

Then $$x = a + 1$$, $$y = b + 3$$, $$z = c + 4$$, and:

$$(a + 1) + (b + 3) + (c + 4) = 21$$

$$a + b + c = 13$$

Apply stars and bars.

The number of non-negative integer solutions to $$a + b + c = 13$$ is:

$$\binom{13 + 2}{2} = \binom{15}{2} = \frac{15 \times 14}{2} = 105$$

The correct answer is 105.

We need to find the number of non-empty subsets of $$S = \{1, 2, 3, 5, 7, 10, 11\}$$ whose element sum is a multiple of 3.

• Residue 0: $$\{3\}$$ — 1 element
• Residue 1: $$\{1, 7, 10\}$$ — 3 elements
• Residue 2: $$\{2, 5, 11\}$$ — 3 elements

Residue 0 group $$\{3\}$$: Including or excluding 3 does not affect the sum mod 3. So this element provides a factor of 2 (include or exclude).

Residue 1 group $$\{1, 7, 10\}$$: Each subset of this group has a sum with some residue mod 3. There are $$2^3 = 8$$ subsets (including empty).

• Sum $$\equiv 0$$: $$\emptyset$$ (sum 0), $$\{1, 7, 10\}$$ (sum 18) — 2 subsets
• Sum $$\equiv 1$$: $$\{1\}$$ (1), $$\{7\}$$ (7), $$\{10\}$$ (10) — 3 subsets
• Sum $$\equiv 2$$: $$\{1,7\}$$ (8), $$\{1,10\}$$ (11), $$\{7,10\}$$ (17) — 3 subsets

Residue 2 group $$\{2, 5, 11\}$$: Similarly, $$2^3 = 8$$ subsets.

• Sum $$\equiv 0$$: $$\emptyset$$ (sum 0), $$\{2, 5, 11\}$$ (sum 18) — 2 subsets
• Sum $$\equiv 1$$: $$\{2, 5\}$$ (7), $$\{2, 11\}$$ (13), $$\{5, 11\}$$ (16) — 3 subsets
• Sum $$\equiv 2$$: $$\{2\}$$ (2), $$\{5\}$$ (5), $$\{11\}$$ (11) — 3 subsets

For the total sum to be $$\equiv 0 \pmod{3}$$, the contributions from residue-1 and residue-2 groups must satisfy:

$$r_1 + r_2 \equiv 0 \pmod{3}$$

Valid combinations $$(r_1, r_2)$$:

• $$(0, 0)$$: $$2 \times 2 = 4$$ subsets
• $$(1, 2)$$: $$3 \times 3 = 9$$ subsets
• $$(2, 1)$$: $$3 \times 3 = 9$$ subsets

Total from residue 1 and 2 groups = $$4 + 9 + 9 = 22$$ subsets.

Element 3 can be included or excluded without affecting divisibility by 3:

$$22 \times 2 = 44 \text{ subsets (including empty set)}$$

$$44 - 1 = 43$$

The answer is $$43$$.

We need to find the total number of six-digit numbers formed using the digits $$4, 5, 9$$ only, that are divisible by $$6$$.

A number is divisible by $$6$$ if it is divisible by both $$2$$ and $$3$$. First, for divisibility by $$2$$, the last digit must be even. Among $$\{4, 5, 9\}$$, only $$4$$ is even, so the last digit must be $$4$$. Since the last digit is fixed as $$4$$, divisibility by $$3$$ requires the sum of the first five digits plus $$4$$ to satisfy $$\text{Sum of first 5 digits} + 4 \equiv 0 \pmod{3}$$, which means $$\text{Sum of first 5 digits} \equiv 2 \pmod{3}$$.

Each of the five positions can be filled with $$4$$, $$5$$, or $$9$$, whose residues modulo $$3$$ are: $$4 \equiv 1 \pmod{3}$$, $$5 \equiv 2 \pmod{3}$$, and $$9 \equiv 0 \pmod{3}$$. Since the residues $$\{0, 1, 2\}$$ are equally likely and each position independently takes any residue, the number of 5-tuples with sum $$\equiv r \pmod{3}$$ is the same for each $$r \in \{0, 1, 2\}$$. The total number of 5-tuples is $$3^5 = 243$$, so there are $$\frac{243}{3} = 81$$ 5-tuples for each residue class.

We need the sum of the first five digits to be $$\equiv 2 \pmod{3}$$, which gives $$81$$ valid combinations for those digits. Therefore, the total number of six-digit numbers is 81.

We need to find the number of ways to select 5 fruits from 7 red apples, 5 white apples, and 8 oranges, with at least 2 oranges, at least 1 red apple, and at least 1 white apple.

Let $$r$$, $$w$$, and $$o$$ be the number of red apples, white apples, and oranges respectively. We need:

• $$r + w + o = 5$$
• $$o \geq 2$$, $$r \geq 1$$, $$w \geq 1$$
• $$r \leq 7$$, $$w \leq 5$$, $$o \leq 8$$

The possible cases are:

Case 1: $$o = 2$$, $$r + w = 3$$ with $$r \geq 1, w \geq 1$$.

Possible: $$(r, w) \in \{(1, 2), (2, 1)\}$$

• $$(1, 2, 2)$$: $$\binom{7}{1} \cdot \binom{5}{2} \cdot \binom{8}{2} = 7 \times 10 \times 28 = 1960$$
• $$(2, 1, 2)$$: $$\binom{7}{2} \cdot \binom{5}{1} \cdot \binom{8}{2} = 21 \times 5 \times 28 = 2940$$

Case 2: $$o = 3$$, $$r + w = 2$$ with $$r \geq 1, w \geq 1$$.

Only $$(r, w) = (1, 1)$$:

• $$(1, 1, 3)$$: $$\binom{7}{1} \cdot \binom{5}{1} \cdot \binom{8}{3} = 7 \times 5 \times 56 = 1960$$

$$1960 + 2940 + 1960 = 6860$$

The correct answer is $$6860$$.

$$a + b = c + d = S \text{ with distinct digits from } \{0,1,2,3,4,5,6,7\} \text{ containing } 7$$

$$(7, x) \text{ and } (y, z) \text{ where } 7+x = y+z$$

$$x=0, S=7 \implies (y,z) \in \{(1,6), (2,5), (3,4)\} \quad \text{(3 sets)}$$

$$x=1, S=8 \implies (y,z) \in \{(2,6), (3,5)\} \quad \text{(2 sets)}$$

$$x=2, S=9 \implies (y,z) \in \{(3,6), (4,5)\} \quad \text{(2 sets)}$$

$$x=3, S=10 \implies (y,z) \in \{(4,6)\} \quad \text{(1 set)}$$

$$x=4, S=11 \implies (y,z) \in \{(5,6)\} \quad \text{(1 set)}$$

$$\text{Total valid sets of 4 digits} = 3 + 2 + 2 + 1 + 1 = 9$$

$$\text{Arrangements per set} = 2 \text{ (pair swap)} \times 2! \text{ (first pair)} \times 2! \text{ (second pair)} = 8$$

$$\text{Maximum trials} = 9 \times 8 = 72$$

We consider six-digit numbers $$x_1x_2x_3x_4x_5x_6$$ with $$0 < x_1 < x_2 < x_3 < x_4 < x_5 < x_6$$ arranged in increasing order, and we wish to find the sum of digits of the 72nd number.

Since $$x_1>0$$, the digits are chosen from $$\{1, 2, \ldots, 9\}$$ and all six must be strictly increasing, giving a total of $$\binom{9}{6} = 84$$ such numbers. Counting by the value of $$x_1$$ shows that there are $$\binom{8}{5}=56$$ numbers starting with $$x_1=1$$, $$\binom{7}{5}=21$$ starting with $$x_1=2$$, $$\binom{6}{5}=6$$ starting with $$x_1=3$$, and $$\binom{5}{5}=1$$ starting with $$x_1=4$$.

Therefore, the first 56 numbers have $$x_1=1$$, so the numbers in positions 57 through 77 have $$x_1=2$$. Hence the 72nd number is the $$72-56=16$$th number among those with $$x_1=2$$.

Within this group, if $$x_2=3$$ there are $$\binom{6}{4}=15$$ numbers obtained by choosing the remaining four from $$\{4,\ldots,9\}$$, and if $$x_2=4$$ there are $$\binom{5}{4}=5$$ numbers from $$\{5,\ldots,9\}$$. Since 16 exceeds 15, the 16th such number has $$x_2=4$$ and is the first in this subgroup, namely $$2,4,5,6,7,8$$.

Therefore, the 72nd number is $$245678$$, and the sum of its digits is $$2+4+5+6+7+8=32$$.

5-digit numbers using digits {0, 2, 3, 4, 7, 9} with repetition, arranged in ascending order. Find the serial number of 42923.

The digits in order: 0, 2, 3, 4, 7, 9. (6 digits total)

Numbers starting with digits less than 4 (i.e., 2 or 3):

(Can't start with 0 for 5-digit number)

Starting with 2: remaining 4 digits each from {0,2,3,4,7,9} = $$6^4 = 1296$$

Starting with 3: $$6^4 = 1296$$

Total: $$2 \times 1296 = 2592$$

Numbers starting with 4, second digit less than 2:

Second digit = 0: $$6^3 = 216$$

Numbers starting with 42, third digit less than 9:

Third digit $$\in \{0, 2, 3, 4, 7\}$$: $$5 \times 6^2 = 5 \times 36 = 180$$

Numbers starting with 429, fourth digit less than 2:

Fourth digit = 0: $$6^1 = 6$$

Numbers starting with 4292, fifth digit less than 3:

Fifth digit $$\in \{0, 2\}$$: 2 numbers

Total numbers less than 42923:

$$2592 + 216 + 180 + 6 + 2 = 2996$$

Serial number of 42923 = $$2996 + 1 = 2997$$.

The answer is $$\boxed{2997}$$.

To find the largest exponent $$n$$ such that $$3^n$$ (reading $$3n$$ as $$3^n$$ based on the context of such factorial problems) divides $$66!$$, we use Legendre's Formula.

Legendre's Formula

The exponent of a prime $$p$$ in $$m!$$ is given by:

$$E_p(m!) = \left\lfloor \frac{m}{p} \right\rfloor + \left\lfloor \frac{m}{p^2} \right\rfloor + \left\lfloor \frac{m}{p^3} \right\rfloor + \dots$$

Calculation for $$p=3$$ and $$m=66$$

1. $$\lfloor 66/3 \rfloor = 22$$
2. $$\lfloor 66/9 \rfloor = 7$$
3. $$\lfloor 66/27 \rfloor = 2$$
4. $$\lfloor 66/81 \rfloor = 0$$ (Stop here)

Sum: $$22 + 7 + 2 = 31$$

We will use this formula and it is used whenever we are solving these types of questions.

$$^{n-1}C_{r-1}=\frac{\left(n-1\right)!}{\left(n-r\right)!\left(r-1\right)!}=\frac{\left(20-1\right)!}{\left(17\right)!\left(3-1\right)!}=\frac{19\times\ 18}{2\times\ 1}=171$$

The word ASSASSINATION has 13 letters: A(3), S(4), I(2), N(2), T(1), O(1). The vowels are A, A, A, I, I, O (6 vowels), and the consonants are S, S, S, S, N, N, T (7 consonants). Since the vowels must occur together, we treat all 6 vowels as a single block.

We then arrange this block along with the 7 consonants, giving 8 items in total with repetitions S(4), N(2), and T(1). The number of arrangements is $$ \text{Arrangements} = \frac{8!}{4! \times 2! \times 1!} = \frac{40320}{48} = 840 $$. Next, arranging the vowels within their block, where there are A(3), I(2), and O(1), yields $$ \text{Arrangements} = \frac{6!}{3! \times 2! \times 1!} = \frac{720}{12} = 60 $$. Therefore, the total number of arrangements is $$840 \times 60 = 50400$$.

We have to form four-digit numbers with the multiset of digits {2, 2, 1, 3}.
Because two 2’s are identical, the number of distinct arrangements is $$\frac{4!}{2!}=12$$.

Let us compute the required sum by analysing the contribution of each digit in each place (thousands, hundreds, tens, ones).

Step 1: How many numbers have a given digit in a fixed place?
Fix one position, say the thousands place.
• If the digit there is 1, the remaining digits are {2, 2, 3}. Distinct permutations of these three digits $$=\frac{3!}{2!}=3$$.
• If the digit there is 3, the remaining digits are {2, 2, 1}. Again $$3$$ permutations.
• If the digit there is 2, after placing one 2 we are left with {2, 1, 3} (all distinct). Permutations of three distinct digits $$=3!=6$$.
So, in one fixed place, 1 appears 3 times, 3 appears 3 times and 2 appears 6 times.

Step 2: Contribution to the sum from each place
Thousands place (weight $$1000$$):
$$1\cdot3\cdot1000 + 3\cdot3\cdot1000 + 2\cdot6\cdot1000 = 3000 + 9000 + 12000 = 24000$$.

Because every place is symmetrical, the same frequency counts apply to the hundreds, tens and ones places.
• Hundreds place (weight $$100$$): $$1\cdot3\cdot100 + 3\cdot3\cdot100 + 2\cdot6\cdot100 = 300 + 900 + 1200 = 2400$$.
• Tens place (weight $$10$$): $$1\cdot3\cdot10 + 3\cdot3\cdot10 + 2\cdot6\cdot10 = 30 + 90 + 120 = 240$$.
• Ones place (weight $$1$$): $$1\cdot3\cdot1 + 3\cdot3\cdot1 + 2\cdot6\cdot1 = 3 + 9 + 12 = 24$$.

Step 3: Total sum
Add the contributions from all four places:
$$24000 + 2400 + 240 + 24 = 26664$$.

Hence, the sum of all distinct four-digit numbers that can be formed with the digits 2, 1, 2, 3 is $$\mathbf{26664}$$.

A 3-digit number using digits {1,2,3,4,5} with repetition, divisible by 6.

Divisible by 2: last digit must be 2 or 4.

Divisible by 3: sum of digits divisible by 3.

Residues mod 3 of {1,2,3,4,5}: 1→1, 2→2, 3→0, 4→1, 5→2. Count: 0→1, 1→2, 2→2.

Last digit = 2 (residue 2): need $$d_1 + d_2 \equiv 1 \pmod{3}$$.

Pairs: (0,1): 1×2=2, (1,0): 2×1=2, (2,2): 2×2=4. Total = 8.

Last digit = 4 (residue 1): need $$d_1 + d_2 \equiv 2 \pmod{3}$$.

Pairs: (0,2): 1×2=2, (2,0): 2×1=2, (1,1): 2×2=4. Total = 8.

Grand total = $$\mathbf{16}$$.

12 courses (5 language, 7 non-language), choose 5 with at most 2 language.

0 language: $$\binom{7}{5} = 21$$. 1 language: $$\binom{5}{1}\binom{7}{4} = 175$$. 2 language: $$\binom{5}{2}\binom{7}{3} = 350$$.

Total = 21 + 175 + 350 = 546.

A boy needs to select 5 courses from 12, of which 5 are language courses. He can choose at most 2 language courses.

The remaining 12 - 5 = 7 courses are non-language courses.

Case 1: 0 language courses.

$$\binom{7}{5} = 21$$

Case 2: 1 language course.

$$\binom{5}{1} \times \binom{7}{4} = 5 \times 35 = 175$$

Case 3: 2 language courses.

$$\binom{5}{2} \times \binom{7}{3} = 10 \times 35 = 350$$

Total = 21 + 175 + 350 = 546

The answer is 546.

Five-digit numbers formed using digits $$\{1, 2, 3, 5, 7\}$$ with repetitions are arranged in descending order, and we wish to find the position of 35337 in this list. Since the available digits in decreasing order are $$7, 5, 3, 2, 1$$, we first count how many numbers exceed 35337.

Numbers starting with 7 or 5 (first digit $$> 3$$): since there are 2 choices for the first digit and each of the remaining four digits can be any of the five available, we have $$2 \times 5^4 = 2 \times 625 = 1250$$ such numbers.
Next, numbers with first digit 3 and second digit $$> 5$$ (i.e., 7) give $$1 \times 5^3 = 125$$.
Then, numbers with first two digits 35 and third digit $$> 3$$ (i.e., 5 or 7) yield $$2 \times 5^2 = 50$$.

Subsequently, numbers with first three digits 353 and fourth digit $$> 3$$ (i.e., 5 or 7) yield $$2 \times 5 = 10$$.
Furthermore, numbers with first four digits 3533 and fifth digit $$> 7$$ contribute 0 since no digit exceeds 7.

From the above, therefore the serial number of 35337 is given by adding these counts plus 1 for the number itself:
$$1250 + 125 + 50 + 10 + 0 + 1 = 1436$$.
Hence, the serial number of 35337 is 1436.

We need to find the number of derangements of 5 elements -- the number of permutations where no student sits in their allotted seat.

Define the problem formally.

A derangement of $$n$$ elements is a permutation $$\sigma$$ such that $$\sigma(i) \neq i$$ for all $$i$$. In this problem, $$n = 5$$ students, and we need no student in their correct seat.

State the derangement formula.

The number of derangements of $$n$$ elements is given by:

$$ D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!} $$

This formula comes from the inclusion-exclusion principle. If $$A_i$$ is the set of permutations where element $$i$$ is fixed, then:

$$ D_n = |S| - |A_1 \cup A_2 \cup \cdots \cup A_n| $$

By inclusion-exclusion, $$|A_{i_1} \cap \cdots \cap A_{i_k}| = (n-k)!$$ (fix $$k$$ elements, permute the rest), and there are $$\binom{n}{k}$$ ways to choose the $$k$$ elements, giving the formula above.

Calculate $$D_5$$.

$$ D_5 = 5! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right) $$

$$ = 120 \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right) $$

Computing the sum inside the parentheses with a common denominator of 120:

$$ = 120 \left(\frac{120 - 120 + 60 - 20 + 5 - 1}{120}\right) $$

$$ = 120 \times \frac{44}{120} = 44 $$

The number of ways is 44.

For a 4-digit number to be divisible by 15, it must satisfy both of the following conditions simultaneously:

1. Divisibility by 5:

The number must end in 5 (since 0 is not available).

2. Divisibility by 3:

The sum of all four digits must be divisible by 3.

Hence, fix the last digit as 5 and choose the remaining three digits such that the total sum is divisible by 3.

Possible valid selections are:

Group 1: Sets with repeated 1s

$$\{1,1,2\}$$

Sum of digits:

$$1+1+2+5 = 9$$

Since 9 is divisible by 3, the number is divisible by 15.

Number of arrangements of $$1,1,2$$:

$$\frac{3!}{2!} = 3$$

Possible arrangements:

$$(1,1,2), (1,2,1), (2,1,1)$$

Total numbers: $$3$$

--------------------------------------------------

$$\{1,1,5\}$$

Sum of digits:

$$1+1+5+5 = 12$$

Since 12 is divisible by 3, the number is divisible by 15.

Number of arrangements:

$$\frac{3!}{2!} = 3$$

Possible arrangements:

$$(1,1,5), (1,5,1), (5,1,1)$$

Total numbers: $$3$$

--------------------------------------------------

Group 2: Sets with repeated 2s or 5s

$$\{2,2,3\}$$

Sum of digits:

$$2+2+3+5 = 12$$

Valid since 12 is divisible by 3.

Number of arrangements:

$$\frac{3!}{2!} = 3$$

Possible arrangements:

$$(2,2,3), (2,3,2), (3,2,2)$$

Total numbers: $$3$$

--------------------------------------------------

$$\{5,5,3\}$$

Sum of digits:

$$5+5+3+5 = 18$$

Valid since 18 is divisible by 3.

Number of arrangements:

$$\frac{3!}{2!} = 3$$

Possible arrangements:

$$(5,5,3), (5,3,5), (3,5,5)$$

Total numbers: $$3$$

--------------------------------------------------

Group 3: Sets with repeated 3s

$$\{3,3,1\}$$

Sum of digits:

$$3+3+1+5 = 12$$

Valid since 12 is divisible by 3.

Number of arrangements:

$$\frac{3!}{2!} = 3$$

Possible arrangements:

$$(3,3,1), (3,1,3), (1,3,3)$$

Total numbers: $$3$$

--------------------------------------------------

Group 4: Sets with distinct digits

$$\{2,3,5\}$$

Sum of digits:

$$2+3+5+5 = 15$$

Valid since 15 is divisible by 3.

Since all digits are distinct, the number of arrangements is:

$$3! = 6$$

Possible arrangements:

$$(2,3,5), (2,5,3), (3,2,5), (3,5,2), (5,2,3), (5,3,2)$$

Total numbers: $$6$$

--------------------------------------------------

Total number of 4-digit numbers divisible by 15:

$$3+3+3+3+3+6 = 21$$

Hence, the required number of 4-digit numbers is

$$\boxed{21}$$

We need to form 4-letter words from the letters of UNIVERSE with exactly 2 vowels and 2 consonants, without repetition.

First, note that the vowels are U, I, E, E (with E appearing twice), while the consonants are N, V, R, S and are all distinct.

When selecting 2 vowels, there are two possibilities. Both vowels could be distinct, which gives $$\binom{3}{2} = 3$$ ways to choose from {U, I, E}.

Choosing 2 consonants from the set {N, V, R, S} can be done in $$\binom{4}{2} = 6$$ ways.

After selecting the letters, we arrange the 4 chosen letters. If all four letters are distinct, there are $$4! = 24$$ arrangements.

Combining these cases leads to the total count:

$$\text{Total} = 3 \times 6 \times 24 = 432$$

The answer is $$432$$.

We need the number of permutations of 1-7 that contain neither "153" nor "2467" as substrings. The total number of permutations is $$7! = 5040$$.

Counting those containing "153", we treat "153" as a single block, giving 5 objects to arrange: [153], 2, 4, 6, 7, so $$|A| = 5! = 120$$. Counting those containing "2467", we treat "2467" as a single block, giving 4 objects: 1, [2467], 3, 5, so $$|B| = 4! = 24$$.

When both "153" and "2467" are present, the two blocks use all 7 digits, and arranging the blocks [153] and [2467] gives $$|A \cap B| = 2! = 2$$. By inclusion-exclusion, $$|A \cup B| = 120 + 24 - 2 = 142$$, so the number of permutations without these substrings is $$5040 - 142 = 4898$$.

The answer is 4898.

We need to count seven-digit positive integers using digits $$\{1, 2, 3, 4\}$$ with digit sum equal to 12.

Let the digits be $$d_1, d_2, \ldots, d_7$$ where each $$d_i \in \{1, 2, 3, 4\}$$ and $$\sum d_i = 12$$.

Substitute $$e_i = d_i - 1$$, so $$e_i \in \{0, 1, 2, 3\}$$ and:

$$\sum_{i=1}^{7} e_i = 12 - 7 = 5$$

We count non-negative integer solutions to $$e_1 + e_2 + \cdots + e_7 = 5$$ with $$0 \leq e_i \leq 3$$ using inclusion-exclusion.

Without upper bound: $$\binom{5+6}{6} = \binom{11}{6} = 462$$

Subtract cases where at least one $$e_i \geq 4$$: Set $$f_i = e_i - 4 \geq 0$$. The remaining variables sum to $$5 - 4 = 1$$. Number of solutions: $$\binom{1+6}{6} = 7$$. There are $$\binom{7}{1} = 7$$ choices for which variable, so we subtract $$7 \times 7 = 49$$.

Cases where two variables $$\geq 4$$: This requires sum $$\geq 8 > 5$$, which is impossible. So no correction needed.

By inclusion-exclusion:

$$462 - 49 = 413$$

The answer is $$\boxed{413}$$.

We need to find the number of seven-digit odd numbers that can be formed using all the digits 1, 2, 2, 2, 3, 3, 5.

A number is odd if its last digit is odd. The odd digits available here are 1, 3, and 5.

If the last digit is 1, the remaining digits are 2, 2, 2, 3, 3, 5. The number of arrangements of these six digits is $$\dfrac{6!}{3! \cdot 2!} = \dfrac{720}{12} = 60$$.

If the last digit is 3, the remaining digits are 1, 2, 2, 2, 3, 5. In this case there are three 2’s and one each of 1, 3, and 5, so the number of ways to arrange them is $$\dfrac{6!}{3!} = \dfrac{720}{6} = 120$$.

If the last digit is 5, the remaining digits are 1, 2, 2, 2, 3, 3. With three 2’s, two 3’s, and one 1, the count is $$\dfrac{6!}{3! \cdot 2!} = \dfrac{720}{12} = 60$$.

Adding these cases gives $$60 + 120 + 60 = 240$$. Therefore, the total number of such seven-digit odd numbers is $$\boxed{240}$$.

Find the total number of 4-digit numbers whose GCD with 54 is 2.

$$54 = 2 \times 3^3$$

$$\gcd(n, 54) = 2$$ means:

- $$n$$ is divisible by 2 (even), and

- $$\gcd(n, 27) = 1$$, i.e., $$n$$ is not divisible by 3.

4-digit numbers range from 1000 to 9999. Total = 9000.

Even 4-digit numbers: $$\frac{9000}{2} = 4500$$

Multiples of 6 from 1000 to 9999: $$\lfloor 9999/6 \rfloor - \lfloor 999/6 \rfloor = 1666 - 166 = 1500$$

Count = $$4500 - 1500 = 3000$$

The answer is $$\boxed{3000}$$.

We need to count 4-digit numbers $$\leq 2800$$ divisible by 3 or by 11.

4-digit numbers range from 1000 to 2800.

Using inclusion-exclusion: $$|A \cup B| = |A| + |B| - |A \cap B|$$

Divisible by 3 (set A):

From 1000 to 2800: First multiple of 3 $$\geq 1000$$ is 1002. Last $$\leq 2800$$ is 2799.

Count = $$\frac{2799 - 1002}{3} + 1 = \frac{1797}{3} + 1 = 599 + 1 = 600$$

Divisible by 11 (set B):

First multiple of 11 $$\geq 1000$$: $$\lceil 1000/11 \rceil = 91$$, so $$91 \times 11 = 1001$$.

Last $$\leq 2800$$: $$\lfloor 2800/11 \rfloor = 254$$, so $$254 \times 11 = 2794$$.

Count = $$254 - 91 + 1 = 164$$

Divisible by 33 (set A ∩ B):

First $$\geq 1000$$: $$\lceil 1000/33 \rceil = 31$$, so $$31 \times 33 = 1023$$.

Last $$\leq 2800$$: $$\lfloor 2800/33 \rfloor = 84$$, so $$84 \times 33 = 2772$$.

Count = $$84 - 31 + 1 = 54$$

Result:

$$|A \cup B| = 600 + 164 - 54 = 710$$

The answer is $$\boxed{710}$$.

We need to count 3-digit numbers (100-999) divisible by 2 or 3 but not by 7.

First, we count numbers divisible by 2 or 3.

Divisible by 2: $$\lfloor 999/2 \rfloor - \lfloor 99/2 \rfloor = 499 - 49 = 450$$

Divisible by 3: $$\lfloor 999/3 \rfloor - \lfloor 99/3 \rfloor = 333 - 33 = 300$$

Divisible by 6 (both 2 and 3): $$\lfloor 999/6 \rfloor - \lfloor 99/6 \rfloor = 166 - 16 = 150$$

Divisible by 2 or 3: $$450 + 300 - 150 = 600$$

Next, we subtract those also divisible by 7.

We need numbers divisible by (2 or 3) AND 7 = divisible by 14 or 21.

Divisible by 14: $$\lfloor 999/14 \rfloor - \lfloor 99/14 \rfloor = 71 - 7 = 64$$

Divisible by 21: $$\lfloor 999/21 \rfloor - \lfloor 99/21 \rfloor = 47 - 4 = 43$$

Divisible by 42 (LCM of 14 and 21): $$\lfloor 999/42 \rfloor - \lfloor 99/42 \rfloor = 23 - 2 = 21$$

Divisible by 14 or 21: $$64 + 43 - 21 = 86$$

Final answer:

$$ 600 - 86 = 514 $$

Therefore, the number of 3-digit numbers is 514.

We need to find the number of 9-digit numbers formed using all digits of 123412341, such that even digits occupy only even places.

The number 123412341 has 9 digits. Counting each digit:

- Digit 1 appears 3 times

- Digit 2 appears 2 times

- Digit 3 appears 2 times

- Digit 4 appears 2 times

Even digits: 2, 2, 4, 4 (4 even digits total)

Odd digits: 1, 1, 1, 3, 3 (5 odd digits total)

In a 9-digit number with positions numbered 1 through 9:

Even places: 2nd, 4th, 6th, 8th (4 even places)

Odd places: 1st, 3rd, 5th, 7th, 9th (5 odd places)

The condition states that even digits must occupy only even places. Since we have exactly 4 even digits and 4 even places, all even places must be filled with even digits, and all odd places must be filled with odd digits.

We need to arrange {2, 2, 4, 4} in 4 positions. The number of distinct arrangements of these 4 items (with repetitions) is:

We need to arrange {1, 1, 1, 3, 3} in 5 positions. The number of distinct arrangements is:

Since the arrangements in even places and odd places are independent, the total number of valid 9-digit numbers is:

The answer is 60.

We need to find the number of 9-digit numbers formed using all digits of 123412341, where even digits occupy only even places.

We start by identifying the digits in the number 123412341. The digits are: 1, 2, 3, 4, 1, 2, 3, 4, 1.

Here, 1 appears 3 times, 2 appears 2 times, 3 appears 2 times, and 4 appears 2 times. Thus, the even digits are 2, 2, 4, 4 (4 even digits) and the odd digits are 1, 1, 1, 3, 3 (5 odd digits).

Next, we note that in a 9-digit number, the even places are positions 2, 4, 6, and 8. Since there are exactly 4 even digits and 4 even places, all even digits must occupy the even positions, and the odd digits occupy the odd positions.

We then arrange the even digits 2, 2, 4, 4 in the 4 even places. The number of arrangements is $$\frac{4!}{2! \times 2!} = 6$$.

Next, we arrange the odd digits 1, 1, 1, 3, 3 in the 5 odd places (positions 1, 3, 5, 7, 9). The number of arrangements is $$\frac{5!}{3! \times 2!} = 10$$.

Multiplying these results gives the total number of such 9-digit numbers as $$6 \times 10 = 60$$.

Therefore, the total number of 9-digit numbers that can be formed under the given conditions is 60.

Step 1: Understand the conditions on the function

We are given the set $$S = \{1, 2, 3, 4, 5, 6\}$$ which contains 6 elements.

We need to find the number of one-one functions $$f: S \to P(S)$$ such that if $$n < m$$, then $$f(n) \subset f(m)$$.

This implies that the images of the elements under $$f$$ form a strictly increasing chain of nested subsets:

$$f(1) \subset f(2) \subset f(3) \subset f(4) \subset f(5) \subset f(6)$$

Step 2: Analyze the sizes of the subsets

Let $$s_i = |f(i)|$$ denote the number of elements in the set $$f(i)$$. Since the subsets are strictly contained within one another, their sizes must be strictly increasing integers:

$$0 \le s_1 < s_2 < s_3 < s_4 < s_5 < s_6 \le 6$$

Thus, the sequence of sizes $$(s_1, s_2, s_3, s_4, s_5, s_6)$$ is formed by choosing 6 distinct integers from the 7 possible values in the set $$\{0, 1, 2, 3, 4, 5, 6\}$$. This means exactly one size will be missing from the sequence.

Step 3: Count the number of valid chains for each case of the missing size

Case 1: The missing size is 0

The sizes of the sets must be $$(1, 2, 3, 4, 5, 6)$$.

To form this chain, we start with 1 element for $$f(1)$$, add 1 element to get $$f(2)$$, add 1 element to get $$f(3)$$, and so on.

The number of ways to choose which elements are added at each step is equal to the number of permutations of the 6 elements:

$$\text{Ways} = 6! = 720$$

Case 2: The missing size is 6

The sizes of the sets must be $$(0, 1, 2, 3, 4, 5)$$.

Here, $$f(1) = \emptyset$$. At each subsequent step, exactly 1 element is added until $$f(6)$$ has 5 elements, leaving exactly 1 element out of the final set.

The number of ways to distribute the elements into this sequence of additions is:

$$\text{Ways} = 6! = 720$$

Case 3: The missing size is $$m$$, where $$m \in \{1, 2, 3, 4, 5\}$$

There are 5 possible values for the missing size. For any such choice, the size sequence includes 0 and 6, but skips $$m$$.

This means that at the transition from size $$m-1$$ to size $$m+1$$, we must add exactly 2 elements at once, while all other transitions add exactly 1 element.

The number of ways to arrange the 6 elements of $$S$$ into these specific increments (one group of 2 elements, and four groups of 1 element) is given by the multinomial coefficient:

$$\text{Ways} = \frac{6!}{2! \times 1! \times 1! \times 1! \times 1!} = \frac{720}{2} = 360$$

Since there are 5 such missing sizes, the total number of ways for this case is:

$$\text{Ways} = 5 \times 360 = 1800$$

Step 4: Sum all the possible cases

The total number of such one-one functions is the sum of the possibilities from all cases:

$$\text{Total} = 720 + 720 + 1800 = 3240$$

Conclusion:

The number of one-one functions satisfying the given conditions is 3240.

$$\int_0^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} dx = \frac{16+20\sqrt{2}}{15}$$

$$\frac{1}{\alpha} \int_0^{\alpha} x(\sqrt{x+\alpha}+\sqrt{x}) dx = \frac{16+20\sqrt{2}}{15}$$

$$\frac{1}{\alpha} \left[ \int_0^{\alpha} x\sqrt{x+\alpha} \, dx + \int_0^{\alpha} x^{3/2} \, dx \right] = \frac{16+20\sqrt{2}}{15}$$

Using substitution $$x+\alpha = t$$ for the first integral:

$$\int_0^{\alpha} x\sqrt{x+\alpha} \, dx = \int_{\alpha}^{2\alpha} (t-\alpha)\sqrt{t} \, dt = \left[ \frac{2}{5}t^{5/2} - \frac{2}{3}\alpha t^{3/2} \right]_{\alpha}^{2\alpha} = \alpha^{5/2} \left( \frac{4\sqrt{2}}{15} + \frac{4}{15} \right)$$

Evaluating the second integral: $$\int_0^{\alpha} x^{3/2} \, dx = \left[ \frac{2}{5}x^{5/2} \right]_0^{\alpha} = \frac{6}{15}\alpha^{5/2}$$

$$\frac{1}{\alpha} \cdot \alpha^{5/2} \left( \frac{10 + 4\sqrt{2}}{15} \right) = \frac{16+20\sqrt{2}}{15}$$

$$\alpha^{3/2}(10 + 4\sqrt{2}) = 16 + 20\sqrt{2}$$

$$\alpha^{3/2}(10 + 4\sqrt{2}) = 2\sqrt{2}(10 + 4\sqrt{2}) \implies \alpha^{3/2} = 2^{3/2} \implies \alpha = 2$$

$$\sqrt{x} \phi(x) = \int_{\pi/4}^{x} \left( 4\sqrt{2}\sin t - 3\phi'(t) \right) dt$$

Differentiating both sides with respect to $$x$$:

$$\frac{1}{2\sqrt{x}} \phi(x) + \sqrt{x} \phi'(x) = 4\sqrt{2}\sin x - 3\phi'(x)$$

At $$x = \frac{\pi}{4}$$:

$$\phi\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{\pi/4}} \int_{\pi/4}^{\pi/4} \left( 4\sqrt{2}\sin t - 3\phi'(t) \right) dt = 0$$

$$0 + \sqrt{\frac{\pi}{4}} \phi'\left(\frac{\pi}{4}\right) = 4\sqrt{2}\sin\left(\frac{\pi}{4}\right) - 3\phi'\left(\frac{\pi}{4}\right)$$

$$\frac{\sqrt{\pi}}{2} \phi'\left(\frac{\pi}{4}\right) = 4\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) - 3\phi'\left(\frac{\pi}{4}\right)$$

$$\left(\frac{\sqrt{\pi}}{2} + 3\right) \phi'\left(\frac{\pi}{4}\right) = 4$$

$$\left(\frac{\sqrt{\pi} + 6}{2}\right) \phi'\left(\frac{\pi}{4}\right) = 4 \implies \phi'\left(\frac{\pi}{4}\right) = \frac{8}{6 + \sqrt{\pi}}$$

We need to find the number of ordered triplets $$(p, q, r)$$ of truth values such that $$p \vee (q \wedge (p \vee r)) \Rightarrow (q \vee r)$$ is True.

An implication $$A \Rightarrow B$$ is False only when $$A$$ is True and $$B$$ is False.

When is the consequent False?

$$q \vee r = F$$ requires $$q = F$$ and $$r = F$$.

With $$q = F, r = F$$, evaluate the antecedent:

$$p \vee (F \wedge (p \vee F)) = p \vee (F \wedge p) = p \vee F = p$$

So the implication is False only when $$p = T, q = F, r = F$$.

Total possible triplets = $$2^3 = 8$$. Number where implication is False = 1.

Number where implication is True = $$8 - 1 = 7$$.

The figure shows a square divided by 6 equally-spaced horizontal and 6 equally-spaced vertical lines.
Hence each side has 7 grid points, giving an overall lattice of $$7 \times 7$$ points.

Total number of intersection points
$$= 7 \times 7 = 49.$$
Label these points $$A_1, A_2, \ldots, A_{49}$$.

A pair of points is called “friends” if the two points are directly adjacent either in the same row or in the same column (vertical or horizontal neighbours only).

Step 1 - Count all possible pairs of points
For 49 distinct points, the total number of unordered pairs is
$$\binom{49}{2} \;=\; \frac{49 \times 48}{2} = 1176.$$

Step 2 - Count horizontal friend pairs
• Each row contains 7 points.
• Adjacent pairs within one row: $$7-1 = 6$$ pairs.
• There are 7 such rows.
So the number of horizontal friend pairs is
$$7 \times 6 = 42.$$

Step 3 - Count vertical friend pairs
By symmetry, each of the 7 columns also has 7 points and therefore 6 adjacent vertical pairs. Thus
$$7 \times 6 = 42$$ vertical friend pairs.

Step 4 - Total friend pairs
$$42 + 42 = 84.$$

Step 5 - Probability that a randomly selected pair is a friend pair
$$p = \frac{\text{number of friend pairs}}{\text{total pairs}} = \frac{84}{1176} = \frac{1}{14}.$$

Step 6 - Compute $$7p$$
$$7p = 7 \times \frac{1}{14} = \frac{1}{2} = 0.50.$$

Therefore, the required value of $$7p$$ is 0.50.

Let $$x_i$$ be the number of balls selected from the $$i^{\text{th}}$$ box, $$i = 1,2,3,4$$.
Each box must contribute at least one red and one blue, so $$x_i \ge 2$$ and $$x_i \le 5$$ (since a box contains only 5 balls).
The total number of balls to be chosen is

$$x_1 + x_2 + x_3 + x_4 = 10 \quad -(1)$$

Write $$x_i = 2 + y_i$$, where $$y_i \ge 0$$ and $$y_i \le 3$$. Substituting in $$(1)$$ gives

$$2+ y_1 + 2+ y_2 + 2+ y_3 + 2+ y_4 = 10$$ $$\Longrightarrow y_1 + y_2 + y_3 + y_4 = 2 \quad -(2)$$

Equation $$(2)$$ distributes 2 identical “extra” units among 4 boxes, each box receiving at most 3. Only two types of distributions are possible:

Case 1: One box gets both extras ($$2,0,0,0$$ and its permutations).

Case 2: Two boxes get one extra each ($$1,1,0,0$$ and its permutations).

Translate back to $$x_i$$ values.

Case 1: Pattern $$\{4,2,2,2\}$$ (one box contributes 4 balls, the other three contribute 2 each).

Number of ways to choose which box is the “4” box: $$4$$.

Case 2: Pattern $$\{3,3,2,2\}$$ (two boxes contribute 3 balls each).

Number of ways to choose the two “3” boxes: $${}^{4}C_{2}=6$$.

Next, count the colour-wise selections inside a single box. Let $$(r,b)$$ denote “$$r$$ red and $$b$$ blue” chosen from that box (all balls are distinct).

Possible $$(r,b)$$ pairs and their counts:

• $$(1,1)\;(\text{for }x=2):\;{}^{3}C_{1}\,{}^{2}C_{1}=3\cdot2=6$$ ways ⇒ $$f(2)=6$$.
• $$(1,2)\text{ or }(2,1)\;(\text{for }x=3):\;3\cdot1=3,\;3\cdot2=6$$ ⇒ $$f(3)=3+6=9$$.
• $$(2,2)\text{ or }(3,1)\;(\text{for }x=4):\;3\cdot1=3,\;1\cdot2=2$$ ⇒ $$f(4)=3+2=5$$.
• $$(3,2)\;(\text{for }x=5):\;1\cdot1=1$$ ⇒ $$f(5)=1$$.

The function $$f(x)$$ gives the number of ways to choose balls from one box when that box contributes $$x$$ balls.

Case 1 (\{4,2,2,2\}):

Product of ways for a fixed box pattern: $$f(4)\,f(2)^3 = 5\cdot6^3 = 5\cdot216 = 1080$$.

Including the 4 possible positions of the “4” box:

Total ways $$= 4 \times 1080 = 4320$$.

Case 2 (\{3,3,2,2\}):

Product of ways for a fixed pattern: $$f(3)^2\,f(2)^2 = 9^2 \cdot 6^2 = 81\cdot36 = 2916$$.

Including the $${}^{4}C_{2}=6$$ choices of the two “3” boxes:

Total ways $$= 6 \times 2916 = 17496$$.

Add the two cases:

$$4320 + 17496 = 21816$$

Hence the required number of ways equals $$21816$$.

Option A which is: 21816

We need to find the total number of 5-digit numbers formed using digits $$\{1, 2, 3, 5, 6, 7\}$$ without repetition that are multiples of 6.

A number is divisible by 6 if and only if it is divisible by both 2 and 3. For divisibility by 2 the last digit must be even (2 or 6), and for divisibility by 3 the sum of the digits must be divisible by 3.

We must choose 5 digits out of the 6. The sum of all 6 digits is $$1 + 2 + 3 + 5 + 6 + 7 = 24.$$ If we omit a digit $$d$$, the sum of the remaining digits is $$24 - d$$, and for divisibility by 3 we require $$24 - d \equiv 0 \pmod{3},$$ i.e., $$d \equiv 0 \pmod{3}.$$ From the available digits, this means $$d = 3$$ or $$d = 6.$$

Omitting 3 leaves the digits $$\{1, 2, 5, 6, 7\}$$ with sum $$21$$, which is divisible by 3. For divisibility by 2 the last digit can be 2 or 6 (2 choices), and the remaining 4 positions can be filled by the other 4 digits in $$4! = 24$$ ways, giving a count of $$2 \times 24 = 48.$$

Omitting 6 leaves the digits $$\{1, 2, 3, 5, 7\}$$ with sum $$18$$, which is divisible by 3. The only available even digit is 2 (1 choice for the last digit), and the remaining 4 positions can be filled by the other 4 digits in $$4! = 24$$ ways, giving a count of $$1 \times 24 = 24.$$

Combining these cases yields a total of $$48 + 24 = 72,$$ so the correct answer is Option A: $$72$$.

We need to distribute 30 identical candies among four children $$C_1, C_2, C_3, C_4$$ with constraints:

• $$C_1 \geq 0$$, $$C_4 \geq 0$$
• $$4 \leq C_2 \leq 7$$
• $$2 \leq C_3 \leq 6$$

Substitute to remove lower bounds:

Let $$C_2 = 4 + a$$ where $$0 \leq a \leq 3$$, and $$C_3 = 2 + b$$ where $$0 \leq b \leq 4$$.

Then: $$C_1 + C_4 = 30 - (4 + a) - (2 + b) = 24 - a - b$$

Count the number of non-negative integer solutions for $$C_1 + C_4 = 24 - a - b$$:

The number of solutions is $$24 - a - b + 1 = 25 - a - b$$.

Sum over all valid values of $$a$$ and $$b$$:

$$\text{Total} = \sum_{a=0}^{3} \sum_{b=0}^{4} (25 - a - b)$$

For $$a = 0$$: $$(25 + 24 + 23 + 22 + 21) = 115$$

For $$a = 1$$: $$(24 + 23 + 22 + 21 + 20) = 110$$

For $$a = 2$$: $$(23 + 22 + 21 + 20 + 19) = 105$$

For $$a = 3$$: $$(22 + 21 + 20 + 19 + 18) = 100$$

$$\text{Total} = 115 + 110 + 105 + 100 = 430$$

The correct answer is Option D: $$430$$.

1. Reduce the bases modulo 7

First, divide the base numbers by 7 to find their remainders:

$$2021 = 7 \times 288 + 5 \implies 2021 \equiv 5 \equiv -2 \pmod 7$$

$$2022 = 7 \times 288 + 6 \implies 2022 \equiv 6 \equiv -1 \pmod 7$$

2. Evaluate the first term

Substitute the reduced base into the first term:

$$(2021)^{2022} \equiv (-2)^{2022} \pmod 7$$

Since the exponent is even, the negative sign disappears:

$$(-2)^{2022} = 2^{2022}$$

By Fermat's Little Theorem, since 7 is a prime number:

$$2^6 \equiv 1 \pmod 7$$

Now, divide the exponent 2022 by 6:

$$2022 = 6 \times 337 + 0$$

Therefore:

$$2^{2022} = (2^6)^{337} \equiv 1^{337} \equiv 1 \pmod 7$$

3. Evaluate the second term

Substitute the reduced base into the second term:

$$(2022)^{2021} \equiv (-1)^{2021} \pmod 7$$

Since the exponent 2021 is an odd number, any negative base raised to an odd power remains negative:

$$(-1)^{2021} = -1 \equiv 6 \pmod 7$$

4. Combine the results

Add the two evaluated remainders together:

$$(2021)^{2022} + (2022)^{2021} \equiv 1 + (-1) \pmod 7$$

$$1 - 1 = 0$$

Given,

$$S=\sum_{r=1}^{20}(r^2+1)r!$$

Now,

$$r^2+1=(r+1)^2-2r$$

Hence,

$$S=\sum_{r=1}^{20}\left((r+1)^2-2r\right)r!$$

$$=\sum_{r=1}^{20}(r+1)^2r!-2\sum_{r=1}^{20}rr!$$

Using

$$ (r+1)^2r!=(r+1)(r+1)! $$

and

$$ rr!=(r+1)!-r! $$

we get

$$S=\sum_{r=1}^{20}(r+1)(r+1)!-2\sum_{r=1}^{20}\left((r+1)!-r!\right)$$

Now,

$$ (r+1)(r+1)!=(r+2-1)(r+1)! $$

$$=(r+2)!-(r+1)!$$

Therefore,

$$S=\sum_{r=1}^{20}\left((r+2)!-(r+1)!\right)-2\sum_{r=1}^{20}\left((r+1)!-r!\right)$$

Both sums are telescopic.

Hence,

$$S=(22!-2!)-2(21!-1!)$$

$$=22!-2-2\cdot21!+2$$

$$=22!-2\cdot21!$$

$$=21!(22-2)$$

$$=22!\ -\ 2.21!$$

Hence,

$$\boxed{22!\ -\ 2.21!}$$.

The permitted digit set is $$\{0,\,2,\,3,\,4,\,6,\,7\}$$ (six digits).
Every required integer is 4-digit and must lie in the closed interval $$[2022,\,4482]$$.

Denote the thousands, hundreds, tens and units digits by $$T,H,E,U$$ respectively.
Because $$2022\le T\,H\,E\,U\le 4482$$, the thousands digit $$T$$ can take only the values $$2,3,4$$ (digits $$0,6,7$$ are either too small or would exceed $$4482$$).

The smallest admissible number is $$2022$$, so the last three-digit part $$H\,E\,U$$ must satisfy $$H\,E\,U\ge 022$$.

Total 3-digit strings possible with the allowed digits: $$6^3 = 216$$.

We must exclude those strings for which $$H\,E\,U\lt 022$$, i.e. $$000$$ to $$021$$. Among these first 22 numbers, only the following possess digits solely from the permitted set: $$000,002,003,004,006,007,020$$  —  7 strings in all.

Hence valid numbers when $$T=2$$: $$216-7 = 209$$.

All numbers $$3000$$-$$3999$$ automatically lie inside the interval. Each of $$H,E,U$$ can again be any of the 6 allowed digits, giving $$6^3 = 216$$ valid numbers.

Now the highest permissible integer is $$4482$$, so the last three-digit part must satisfy $$H\,E\,U\le 482$$.

Total possible 3-digit strings: $$6^3 = 216$$.
We count directly how many of these are $$\le 482$$.

The hundreds digit $$H$$ can be $$0,2,3,4$$ (digits $$6,7$$ would already yield $$\ge 600$$).

For $$H=0,2,3$$ every choice of $$E,U$$ works (because the number is $$\le 399$$). That yields $$3\times 6\times 6 = 108$$ numbers.

For $$H=4$$ we need $$4E U \le 482 \Longrightarrow 10E+U\le 82$$.
For each allowed tens digit $$E=0,2,3,4,6,7$$ the inequality $$U\le 82-10E$$ is satisfied by all six allowed units digits (each is at most $$7$$). Thus $$6\times 6 = 36$$ additional numbers.

Total for $$T=4$$: $$108+36 = 144$$.

Adding the three cases:

$$209 + 216 + 144 = 569$$

Therefore, the number of 4-digit integers required is 569.

We wish to count bijective functions $$f:\{1,3,5,7,\ldots,99\}\to\{2,4,6,8,\ldots,100\}$$ satisfying the strict inequality $$f(3)>f(5)>f(7)>\cdots>f(99)\,.$$

Both the domain and the codomain consist of 50 elements, namely $$\{1,3,5,\ldots,99\}$$ and $$\{2,4,6,\ldots,100\}$$ respectively. Since $$f$$ is bijective, every one of the 50 codomain values must be used exactly once.

The requirement that the values assigned to the 49 inputs $$\{3,5,7,\ldots,99\}$$ form a strictly decreasing sequence implies that once we choose which 49 codomain values fill those inputs, there is exactly one way to order them so that $$f(3)>f(5)>\cdots>f(99)\,. $$ The remaining input, 1, can be mapped to any of the 50 codomain values; choosing its image then determines exactly which 49 values remain for the decreasing chain.

Therefore, the total number of such bijections is simply the number of choices for $$f(1)$$, namely:

The answer is Option A.

There are 5 MCQ questions with 3 choices each. Scoring: +3 correct, -2 wrong, 0 unattempted. We need the number of ways to score exactly 5 marks.

Set up equations.

Let $$c$$ = correct, $$w$$ = wrong, $$s$$ = skipped. Then:

$$c + w + s = 5$$

$$3c - 2w = 5$$

Find valid combinations.

From $$3c - 2w = 5$$: testing values of $$c$$:

$$c = 1$$: $$w = -1$$ (invalid)

$$c = 3$$: $$w = 2$$, $$s = 0$$ ✓

$$c = 5$$: $$w = 5$$ (but $$c + w = 10 > 5$$, invalid)

The only valid case is $$c = 3, w = 2, s = 0$$.

Count the number of ways.

Choose which 3 questions are answered correctly: $$\binom{5}{3} = 10$$ ways.

For the remaining 2 wrong answers, each has 2 incorrect choices: $$2^2 = 4$$ ways.

Total = $$10 \times 4 = 40$$

Answer: 40

We have $$S = \{4, 6, 9\}$$ and $$T = \{9, 10, 11, \ldots, 1000\}$$. The set $$A$$ consists of all numbers that can be represented as $$a_1 + a_2 + \cdots + a_k$$ where each $$a_i \in \{4, 6, 9\}$$ and $$k \in \mathbb{N}$$. We need to find the sum of all elements in $$T \setminus A$$, i.e., elements of $$T$$ that cannot be expressed as such sums.

We note that every element of $$A$$ is a non-negative integer combination of 4, 6, and 9 (with at least one term). The elements of $$A$$ starting from the smallest are: 4, 6, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, ... We need to find which numbers from 9 onwards are NOT in $$A$$.

We systematically check small values. A number $$n$$ is in $$A$$ if $$n = 4a + 6b + 9c$$ for non-negative integers $$a, b, c$$ with $$a + b + c \geq 1$$:

$$n = 4$$: $$4 = 4(1)$$. Yes. $$n = 6$$: $$6 = 6(1)$$. Yes. $$n = 8$$: $$8 = 4(2)$$. Yes. $$n = 9$$: $$9 = 9(1)$$. Yes. $$n = 10$$: $$10 = 4+6$$. Yes. $$n = 12$$: $$12 = 4(3)$$ or $$6(2)$$. Yes. $$n = 13$$: $$13 = 4+9$$. Yes. $$n = 14$$: $$14 = 4+4+6$$. Yes. $$n = 15$$: $$15 = 6+9$$. Yes. $$n = 16$$: $$16 = 4(4)$$. Yes. $$n = 17$$: $$17 = 4+4+9$$. Yes. $$n = 18$$: $$18 = 9(2)$$ or $$4(3)+6$$. Yes.

Now, $$n = 19$$: $$19 = 4+6+9$$. Yes. $$n = 20$$: $$20 = 4(5)$$. Yes. For $$n = 21$$: $$21 = 12+9 = 4(3)+9$$. Yes. $$n = 22$$: $$22 = 4+9+9 = 4+18$$. Yes. $$n = 23$$: $$23 = 4+4+6+9 = 14+9$$. Yes.

We observe that all integers $$\geq 24$$ can be written as sums: if $$n \equiv 0 \pmod{4}$$, use $$4$$'s; if $$n \equiv 1 \pmod{4}$$, use $$9 + 4$$'s (since $$n \geq 25$$ gives enough room, and $$n = 9$$ works, $$n = 13$$ works); if $$n \equiv 2 \pmod{4}$$, use $$6 + 4$$'s; if $$n \equiv 3 \pmod{4}$$, use $$9 + 6 + 4$$'s.

More precisely, for $$n \geq 24$$: $$n \equiv 0 \pmod{2}$$ can always be written as $$4a + 6b$$; $$n$$ odd and $$n \geq 9$$ can be written as $$9 + \text{even number} \geq 0$$. Since $$n - 9 \geq 15$$ for $$n \geq 24$$, and $$n - 9$$ is even, and any even number $$\geq 4$$ can be written as $$4a + 6b$$ (and even numbers $$\geq 4$$: 4, 6, 8, 10, ... all work). Actually even number 0 also works (take nothing extra). So for odd $$n \geq 9$$: $$n - 9$$ is even and $$\geq 0$$, and we need $$n - 9 \neq 2$$ (since 2 cannot be made from 4s and 6s). $$n - 9 = 2 \Rightarrow n = 11$$. So $$n = 11$$ is the concern.

Let us check $$n = 11$$: Can we write $$11 = 4a + 6b + 9c$$? With $$c = 1$$: $$2 = 4a + 6b$$, no non-negative solution. With $$c = 0$$: $$11 = 4a + 6b$$, impossible (parity: $$4a + 6b$$ is even, 11 is odd). So $$n = 11 \notin A$$.

We also need to check numbers less than 24 more carefully. From the set $$T = \{9, 10, \ldots, 1000\}$$, which values from 9 to 23 are NOT in $$A$$?

$$n = 9$$: Yes (checked). $$n = 10$$: Yes. $$n = 11$$: No (shown above). $$n = 12$$: Yes. $$n = 13$$: Yes. $$n = 14$$: Yes. $$n = 15$$: Yes. $$n = 16$$: Yes. $$n = 17$$: Yes. $$n = 18$$: Yes. $$n = 19$$: Yes. $$n = 20$$: Yes. $$n = 21$$: Yes. $$n = 22$$: Yes. $$n = 23$$: Yes.

For $$n \geq 24$$, every number is representable as argued. So the only element of $$T$$ not in $$A$$ is $$n = 11$$.

Therefore, the sum of all elements in $$T \setminus A$$ is $$\boxed{11}$$.

Hence, the correct answer is 11.

We need to find how many 4-digit numbers between 1000 and 3000, divisible by 4, can be formed using digits 1, 2, 3, 4, 5, 6 without repetition.

Since the number is between 1000 and 3000, the first digit is either 1 or 2.

A number is divisible by 4 if its last two digits form a number divisible by 4. The possible two-digit endings from {1, 2, 3, 4, 5, 6} (without repetition between the two digits) that are divisible by 4 are:

12, 16, 24, 32, 36, 52, 56, 64

Remaining digits available: {2, 3, 4, 5, 6} (digit 1 is used).

Valid endings not using digit 1: 24, 32, 36, 52, 56, 64 — that gives 6 endings.

For each ending, the second digit can be any of the remaining 3 digits (from the 5 available, minus the 2 used in the ending).

Count: $$6 \times 3 = 18$$

Remaining digits available: {1, 3, 4, 5, 6} (digit 2 is used).

Valid endings not using digit 2: 16, 36, 56, 64 — that gives 4 endings.

(Endings 12, 24, 32, 52 all use digit 2, so they are excluded.)

For each ending, the second digit can be any of the remaining 3 digits.

Count: $$4 \times 3 = 12$$

$$18 + 12 = 30$$

The correct answer is $$\boxed{30}$$.

We need to arrange 16 cubes (11 blue, 5 red) in a row such that between any two red cubes there are at least 2 blue cubes.

First, we place all 11 blue cubes in a row. This creates 12 gaps (including ends): $$\_B\_B\_B\_B\_B\_B\_B\_B\_B\_B\_B\_$$

Since the condition requires at least 2 blue cubes between consecutive red cubes, we can model this using a change of variable.

Let us place the 5 red cubes among the 11 blue cubes. Let $$b_0, b_1, b_2, b_3, b_4, b_5$$ be the number of blue cubes before the first red, between consecutive red cubes, and after the last red respectively.

We need: $$b_0 + b_1 + b_2 + b_3 + b_4 + b_5 = 11$$ with $$b_1, b_2, b_3, b_4 \geq 2$$ and $$b_0, b_5 \geq 0$$.

Substituting to remove these constraints, let $$b_i' = b_i - 2$$ for $$i = 1, 2, 3, 4$$. Then $$b_i' \geq 0$$ and:

$$b_0 + (b_1' + 2) + (b_2' + 2) + (b_3' + 2) + (b_4' + 2) + b_5 = 11$$

$$b_0 + b_1' + b_2' + b_3' + b_4' + b_5 = 11 - 8 = 3$$

Now, the number of non-negative integer solutions to $$b_0 + b_1' + b_2' + b_3' + b_4' + b_5 = 3$$ (6 variables) is:

$$\binom{3 + 6 - 1}{6 - 1} = \binom{8}{5} = 56$$

The answer is $$\boxed{56}$$.

We need to find the total number of four-digit numbers such that each of the first three digits is divisible by the last digit.

Let the four-digit number be $$\overline{abcd}$$ where $$a$$ is the thousands digit, $$b$$ the hundreds, $$c$$ the tens, and $$d$$ the units digit. The conditions are $$d | a$$, $$d | b$$, $$d | c$$, $$a \geq 1$$, and $$d \geq 1$$.

d = 1: Any digit is divisible by 1, so $$a \in \{1,...,9\}$$ (9 choices), $$b, c \in \{0,...,9\}$$ (10 choices each), giving $$9 \times 10 \times 10 = 900$$.

d = 2: $$a \in \{2,4,6,8\}$$ (4 choices), $$b, c \in \{0,2,4,6,8\}$$ (5 choices each), giving $$4 \times 5 \times 5 = 100$$.

d = 3: $$a \in \{3,6,9\}$$ (3 choices), $$b, c \in \{0,3,6,9\}$$ (4 choices each), giving $$3 \times 4 \times 4 = 48$$.

d = 4: $$a \in \{4,8\}$$ (2 choices), $$b, c \in \{0,4,8\}$$ (3 choices each), giving $$2 \times 3 \times 3 = 18$$.

d = 5: $$a \in \{5\}$$ (1 choice), $$b, c \in \{0,5\}$$ (2 choices each), giving $$1 \times 2 \times 2 = 4$$.

d = 6: $$a \in \{6\}$$ (1 choice), $$b, c \in \{0,6\}$$ (2 choices each), giving $$1 \times 2 \times 2 = 4$$.

d = 7: $$a \in \{7\}$$ (1 choice), $$b, c \in \{0,7\}$$ (2 choices each), giving $$1 \times 2 \times 2 = 4$$.

d = 8: $$a \in \{8\}$$ (1 choice), $$b, c \in \{0,8\}$$ (2 choices each), giving $$1 \times 2 \times 2 = 4$$.

d = 9: $$a \in \{9\}$$ (1 choice), $$b, c \in \{0,9\}$$ (2 choices each), giving $$1 \times 2 \times 2 = 4$$.

$$900 + 100 + 48 + 18 + 4 + 4 + 4 + 4 + 4 = 1086$$ Therefore, the answer is $$\textbf{1086}$$.

We need to count three-digit numbers where exactly one digit is repeated exactly two times (i.e., the number has exactly two identical digits and one different digit).

Case 1: Repeated digit $$\neq 0$$ and different digit $$\neq 0$$

Choose the repeated digit: 9 choices (1-9)

Choose the different digit: 8 choices (1-9, excluding the repeated digit)

Choose which position gets the different digit: 3 positions

Count: $$9 \times 8 \times 3 = 216$$

Case 2: Repeated digit $$\neq 0$$ and different digit $$= 0$$

Choose the repeated digit: 9 choices (1-9)

The different digit is 0, which cannot be in the hundreds place

Position for 0: 2 choices (tens or units place)

Count: $$9 \times 1 \times 2 = 18$$

Case 3: Repeated digit $$= 0$$ and different digit $$\neq 0$$

Two positions have digit 0 and one has a non-zero digit

The non-zero digit must be in the hundreds place (since a three-digit number cannot start with 0)

Choose the non-zero digit: 9 choices (1-9)

The non-zero digit must go in the hundreds position: 1 way

Count: $$9 \times 1 = 9$$

Total:

$$216 + 18 + 9 = 243$$

The answer is $$\boxed{243}$$.

We need $$\binom{b}{3} \cdot \binom{g}{2} = 168$$.

We factorize: $$168 = 8 \times 21 = 2^3 \times 3 \times 7$$.

Let us try small values. We need $$\binom{b}{3}$$ and $$\binom{g}{2}$$ to be factors of 168.

$$\binom{b}{3}$$: for $$b = 3$$: 1; $$b = 4$$: 4; $$b = 5$$: 10; $$b = 6$$: 20; $$b = 7$$: 35; $$b = 8$$: 56.

$$\binom{g}{2}$$: for $$g = 2$$: 1; $$g = 3$$: 3; $$g = 4$$: 6; $$g = 5$$: 10; $$g = 6$$: 15; $$g = 7$$: 21.

We need the product to be 168:

If $$\binom{b}{3} = 56$$ ($$b = 8$$), then $$\binom{g}{2} = 3$$ ($$g = 3$$). Check: $$56 \times 3 = 168$$. Yes!

If $$\binom{b}{3} = 4$$ ($$b = 4$$), then $$\binom{g}{2} = 42$$, and $$\binom{g}{2} = 42$$ gives $$g(g-1) = 84$$. Since $$9 \times 8 = 72$$ and $$10 \times 9 = 90$$, no integer solution.

If $$\binom{b}{3} = 35$$ ($$b = 7$$), then $$\binom{g}{2} = 168/35 = 4.8$$. Not integer.

If $$\binom{b}{3} = 20$$ ($$b = 6$$), then $$\binom{g}{2} = 168/20 = 8.4$$. Not integer.

If $$\binom{b}{3} = 10$$ ($$b = 5$$), then $$\binom{g}{2} = 168/10 = 16.8$$. Not integer.

So the only valid solution is $$b = 8, g = 3$$.

Therefore $$b + 3g = 8 + 9 = 17$$.

Hence, the correct answer is 17.

We need to count 4-element permutations $$b_1b_2b_3b_4$$ with $$b_i \in \{1,2,\ldots,100\}$$, all distinct, such that either $$b_1, b_2, b_3$$ are consecutive integers or $$b_2, b_3, b_4$$ are consecutive integers. Here "consecutive integers" means they appear in increasing consecutive order.

For permutations where $$b_1, b_2, b_3$$ are consecutive, we require $$b_2 = b_1 + 1$$ and $$b_3 = b_1 + 2$$. The starting value $$b_1$$ can be any integer from 1 to 98, giving 98 choices, and the fourth element $$b_4$$ can be any of the remaining $$100 - 3 = 97$$ values. Thus $$|A| = 98 \times 97 = 9506$$.

Similarly, for permutations where $$b_2, b_3, b_4$$ are consecutive, we require $$b_3 = b_2 + 1$$ and $$b_4 = b_2 + 2$$. The starting value $$b_2$$ can be any integer from 1 to 98, giving 98 choices, and the first element $$b_1$$ can be any of the remaining $$100 - 3 = 97$$ values. Hence $$|B| = 98 \times 97 = 9506$$.

In the overlap where both conditions hold, we have from the first condition $$b_2 = b_1 + 1$$, $$b_3 = b_1 + 2$$ and from the second condition $$b_3 = b_2 + 1 = b_1 + 2$$ and $$b_4 = b_2 + 2 = b_1 + 3$$. Therefore all four elements are consecutive: $$b_1, b_1+1, b_1+2, b_1+3$$, with the starting value $$b_1$$ ranging from 1 to 97, giving $$|A \cap B| = 97$$.

By inclusion-exclusion, $$|A \cup B| = |A| + |B| - |A \cap B| = 9506 + 9506 - 97 = 18915$$.

The answer is $$\boxed{18915}$$.

Passwords are 6 to 8 characters long. Each character is from {A, B, C, D, E, 1, 2, 3, 4, 5} (10 characters total), and we require at least one number. Therefore, the total number of passwords is $$10^6 + 10^7 + 10^8.$$

To exclude those with no numbers, we consider passwords using only the five letters {A, B, C, D, E}, which gives $$5^6 + 5^7 + 5^8. $$

Subtracting these all-letter passwords from the total yields $$= (10^6 + 10^7 + 10^8) - (5^6 + 5^7 + 5^8). $$

Rewriting the difference term by term, we have $$= (10^6 - 5^6) + (10^7 - 5^7) + (10^8 - 5^8). $$

Since $$10^n = (2 \cdot 5)^n = 2^n \cdot 5^n$$, we can factor each term as follows: $$= 5^6(2^6 - 1) + 5^7(2^7 - 1) + 5^8(2^8 - 1). $$

Grouping the factors, this becomes $$= 5^6[63 + 5 \times 127 + 25 \times 255], $$ so that $$= 5^6[63 + 635 + 6375]. $$

Therefore, we obtain $$= 5^6 \times 7073, $$ and hence $$\alpha = 7073. $$

The answer is 7073.

We need to find the serial number of the word "MANKIND" when all permutations of its letters are arranged in dictionary order.

First, we list the letters in alphabetical order. MANKIND has letters: M, A, N, K, I, N, D. Sorted alphabetically, they become A, D, I, K, M, N, N; note that N appears twice.

Since each word starting with a letter before M (namely A, D, I, K) leaves six letters with two Ns, each such prefix yields $$\dfrac{6!}{2!} = 360$$ arrangements, giving a subtotal of $$4\times 360 = 1440$$ words before those beginning with M.

When the first letter is M, no letter precedes A, so no words begin with MA that come before "MANKIND".

Once MA is fixed, the remaining letters are D, I, K, N, N. Because D, I, and K all come before N, each choice leads to $$\dfrac{4!}{2!} = 12$$ permutations, totaling $$3\times 12 = 36$$ words before MAN.

Fixing MAN leaves D, I, K, N; among these, D and I precede K, and each gives $$3! = 6$$ permutations, so there are $$2\times 6 = 12$$ words before MANK.

With MANK as prefix, the remaining letters are D, I, N, and only D comes before I, giving $$2! = 2$$ words before MANKI.

When the prefix is MANKI, the leftover letters are D and N, and D precedes N, yielding $$1! = 1$$ word before MANKIN.

Finally, for words starting with MANKIN, the only remaining letter is D, so there are no further words before MANKIND.

$$\text{Rank} = 1440 + 36 + 12 + 2 + 1 + 0 + 1 = 1492$$

The +1 in the sum accounts for the word "MANKIND" itself, so it occupies the position $$1492$$ in the dictionary order.

Let the 3-digit odd number be

$$100a+10b+c$$

where

$$a\in\{1,2,\ldots,9\},\qquad b\in\{0,1,\ldots,9\}$$

and since the number is odd,

$$c\in\{1,3,5,7,9\}$$

Also,

$$a+b+c$$

must be a multiple of $$7$$.

Possible multiples of $$7$$ between minimum sum $$2$$ and maximum sum $$27$$ are

$$7,\ 14,\ 21$$

Now, count case-wise.

Case 1: $$c=1$$

Then,

$$a+b=6,13,20$$

For

$$a+b=6$$

number of solutions $$=6$$

For

$$a+b=13$$

number of solutions $$=6$$

For

$$a+b=20$$

number of solutions $$=0$$

Total cases $$=12$$

Case 2: $$c=3$$

Then,

$$a+b=4,11,18$$

For

$$a+b=4$$

number of solutions $$=4$$

For

$$a+b=11$$

number of solutions $$=8$$

For

$$a+b=18$$

number of solutions $$=1$$

Total cases $$=13$$

Case 3: $$c=5$$

Then,

$$a+b=2,9,16$$

For

$$a+b=2$$

number of solutions $$=2$$

For

$$a+b=9$$

number of solutions $$=9$$

For

$$a+b=16$$

number of solutions $$=3$$

Total cases $$=14$$

Case 4: $$c=7$$

Then,

$$a+b=0,7,14,21$$

For

$$a+b=0$$

number of solutions $$=0$$

For

$$a+b=7$$

number of solutions $$=7$$

For

$$a+b=14$$

number of solutions $$=5$$

For

$$a+b=21$$

number of solutions $$=0$$

Total cases $$=12$$

Case 5: $$c=9$$

Then,

$$a+b=5,12,19$$

For

$$a+b=5$$

number of solutions $$=5$$

For

$$a+b=12$$

number of solutions $$=7$$

For

$$a+b=19$$

number of solutions $$=0$$

Total cases $$=12$$

Hence, total number of required numbers

$$=12+13+14+12+12$$

$$=63$$

Therefore, the required number is $$\boxed{63}$$.

We need the number of 5-digit numbers whose product of digits is

$$36$$

Prime factorization:

$$36=2^2\cdot3^2$$

Possible non-zero digit combinations are:

$$36=1\cdot1\cdot1\cdot1\cdot36\quad (\text{not possible})$$

Now list valid digit multisets using digits $$0-9.$$

Case 1:

$$36=1\cdot1\cdot1\cdot4\cdot9$$

Number of arrangements:

$$\frac{5!}{3!}=20$$

Case 2:

$$36=1\cdot1\cdot1\cdot6\cdot6$$

Number of arrangements:

$$\frac{5!}{3!2!}=10$$

Case 3:

$$36=1\cdot1\cdot2\cdot2\cdot9$$

Number of arrangements:

$$\frac{5!}{2!2!}=30$$

Case 4:

$$36=1\cdot1\cdot2\cdot3\cdot6$$

Number of arrangements:

$$\frac{5!}{2!}=60$$

Case 5:

$$36=1\cdot1\cdot3\cdot3\cdot4$$

Number of arrangements:

$$\frac{5!}{2!2!}=30$$

Case 6:

$$36=1\cdot2\cdot2\cdot3\cdot3$$

Number of arrangements:

$$\frac{5!}{2!2!}=30$$

Total numbers:

$$20+10+30+60+30+30=180$$

Hence, the required number of 5-digit numbers is

$$\boxed{180}$$.

We need to find the number of 7-digit numbers formed using all the digits $$1, 2, 3, 4, 5, 7, 9$$ (each used exactly once) that are multiples of 11.

Divisibility rule for 11: A number is divisible by 11 if the alternating sum (sum of digits at odd positions minus sum of digits at even positions) is divisible by 11.

The total sum of all seven digits: $$1 + 2 + 3 + 4 + 5 + 7 + 9 = 31$$

For a 7-digit number $$d_1 d_2 d_3 d_4 d_5 d_6 d_7$$:

Positions 1, 3, 5, 7 (odd positions): 4 digits with sum $$S_o$$

Positions 2, 4, 6 (even positions): 3 digits with sum $$S_e$$

We need: $$S_o + S_e = 31$$ and $$S_o - S_e \equiv 0 \pmod{11}$$

Adding these: $$2S_o = 31 + 11k$$ for some integer $$k$$. Since $$S_o$$ must be an integer, $$31 + 11k$$ must be even, so $$k$$ must be odd.

Case $$k = 1$$: $$S_o = \frac{31+11}{2} = 21$$, $$S_e = 10$$

Case $$k = -1$$: $$S_o = \frac{31-11}{2} = 10$$, $$S_e = 21$$

Case $$k = 3$$: $$S_o = \frac{31+33}{2} = 32$$. The maximum possible $$S_o$$ with 4 digits from our set is $$4 + 5 + 7 + 9 = 25 < 32$$. Not possible.

Case $$k = -3$$: $$S_o = \frac{31-33}{2} = -1 < 0$$. Not possible.

So only $$k = \pm 1$$ are valid.

Finding 4-element subsets of $$\{1,2,3,4,5,7,9\}$$ with sum 21:

We systematically check all combinations including 9:

$$\{9, 7, 3, 2\}$$: $$9+7+3+2 = 21$$ $$\checkmark$$

$$\{9, 7, 4, 1\}$$: $$9+7+4+1 = 21$$ $$\checkmark$$

$$\{9, 5, 4, 3\}$$: $$9+5+4+3 = 21$$ $$\checkmark$$

$$\{9, 7, 5, ??\}$$: Need sum 0 from remaining - not possible.

$$\{9, 5, 3, 4\}$$: Already counted above.

Combinations without 9:

$$\{7, 5, 4, 3\}$$: $$7+5+4+3 = 19 \neq 21$$

$$\{7, 5, 4, 2\}$$: $$= 18$$. No.

Maximum without 9: $$\{7, 5, 4, 3\} = 19 < 21$$. None work.

So there are exactly 3 subsets with $$S_o = 21$$.

Finding 3-element subsets of $$\{1,2,3,4,5,7,9\}$$ with sum 21:

$$\{9, 7, 5\}$$: $$9+7+5 = 21$$ $$\checkmark$$

$$\{9, 7, 4\}$$: $$= 20$$. No.

$$\{9, 7, 3\}$$: $$= 19$$. No.

Maximum of any other triple: $$9+7+4 = 20 < 21$$, and $$9+7+5 = 21$$ is the only one.

So there is exactly 1 subset with $$S_e = 21$$.

Counting arrangements:

For each valid partition of digits into odd and even positions:

The 4 digits at odd positions can be arranged in $$4! = 24$$ ways.

The 3 digits at even positions can be arranged in $$3! = 6$$ ways.

Each subset contributes $$4! \times 3! = 24 \times 6 = 144$$ numbers.

Case 1 ($$S_o = 21$$): 3 subsets $$\times$$ 144 = 432

Case 2 ($$S_o = 10, S_e = 21$$): The 3-element subset $$\{5, 7, 9\}$$ goes to even positions, and the remaining 4 digits $$\{1, 2, 3, 4\}$$ (sum = 10) go to odd positions. This gives 1 subset $$\times$$ 144 = 144.

Total: $$432 + 144 = 576$$

The correct answer is $$576$$.

We need to count natural numbers between 1012 and 23421, formed using the digits $$\{2, 3, 4, 5, 6\}$$ without repetition, that are divisible by 55. Since $$55 = 5 \times 11$$, the number must be divisible by both 5 and 11. The only available digit that is a multiple of 5 is 5 itself (and 0 is not available), so the last digit must be 5.

The range 1012 to 23421 includes 4-digit numbers and 5-digit numbers. Since we use distinct digits from $$\{2,3,4,5,6\}$$, a 4-digit number uses exactly four of these five digits, and a 5-digit number uses all five.

Case 1: 4-digit numbers $$\overline{abc5}$$ where $$a, b, c$$ are three distinct digits from $$\{2, 3, 4, 6\}$$. For divisibility by 11, the alternating sum of a 4-digit number $$\overline{d_1 d_2 d_3 d_4}$$ must satisfy $$(d_1 - d_2 + d_3 - d_4) \equiv 0 \pmod{11}$$. Here this gives $$(a - b + c - 5) \equiv 0 \pmod{11}$$, i.e., $$(a + c) - b = 5 + 11k$$ for integer $$k$$.

Since $$a, b, c \in \{2,3,4,6\}$$ are distinct, $$(a+c) - b$$ ranges from $$(2+3) - 6 = -1$$ to $$(6+4) - 2 = 8$$. The only multiple-of-11 shift that fits is $$k = 0$$, so $$(a+c) - b = 5$$.

Writing $$a + b + c = S$$ and using $$a + c = b + 5$$, we get $$S = 2b + 5$$, so $$b = \frac{S-5}{2}$$.

For the triplet $$\{2,3,4\}$$ with $$S = 9$$: $$b = 2$$, and $$a, c \in \{3,4\}$$. This gives numbers 3245 (since $$3245 = 55 \times 59$$) and 4235 (since $$4235 = 55 \times 77$$). Both are in range.

For $$\{2,3,6\}$$ with $$S = 11$$: $$b = 3$$, and $$a, c \in \{2,6\}$$. This gives numbers 2365 (since $$2365 = 55 \times 43$$) and 6325 (since $$6325 = 55 \times 115$$). Both are in range.

For $$\{2,4,6\}$$ with $$S = 12$$: $$b = 3.5$$, which is not a valid digit. No solutions here.

For $$\{3,4,6\}$$ with $$S = 13$$: $$b = 4$$, and $$a, c \in \{3,6\}$$. This gives numbers 3465 (since $$3465 = 55 \times 63$$) and 6435 (since $$6435 = 55 \times 117$$). Both are in range.

This yields $$2 + 2 + 0 + 2 = 6$$ valid 4-digit numbers.

Case 2: 5-digit numbers using all five digits $$\{2,3,4,5,6\}$$ with last digit 5, lying between 10000 and 23421. The first digit must be 2 (since digits 3, 4, or 6 as the leading digit would produce numbers exceeding 30000). With first digit 2, the number is $$\overline{2\,b\,c\,d\,5}$$ where $$b, c, d$$ is a permutation of $$\{3,4,6\}$$.

The alternating sum is $$2 - b + c - d + 5 = (7 + c) - (b + d)$$. Since $$b + c + d = 13$$, we have $$b + d = 13 - c$$, so the alternating sum equals $$7 + c - 13 + c = 2c - 6$$. For divisibility by 11, we need $$2c - 6 \equiv 0 \pmod{11}$$, giving $$c = 3$$. With $$c = 3$$, $$\{b,d\} = \{4,6\}$$, producing numbers 24365 and 26345. Since both exceed 23421, neither is valid.

The total count is $$6 + 0 = 6$$.

Hence, the correct answer is $$\boxed{6}$$.

Ten boys $$B_1, B_2, \ldots, B_{10}$$ and five girls $$G_1, G_2, \ldots, G_5$$. We need groups of 3 boys and 3 girls where $$B_1$$ and $$B_2$$ are NOT both in the group.

Without any restriction, the total number of ways to select 3 boys from 10 and 3 girls from 5 is $$\binom{10}{3} \times \binom{5}{3} = 120 \times 10 = 1200$$.

However, we must subtract the cases in which both $$B_1$$ and $$B_2$$ are chosen. In such situations, we have already selected two boys, so the remaining boy must be chosen from the other 8 boys, and the girls are still chosen from all 5, giving $$\binom{8}{1} \times \binom{5}{3} = 8 \times 10 = 80$$.

Subtracting these restricted cases from the total yields $$1200 - 80 = 1120$$.

The answer is $$\boxed{1120}$$.

Given,

$$A=\{1,2,3,4,5,6,7\}$$

and

$$B=\{T\subseteq A:\text{ either }1\notin T\text{ or }2\in T\}$$

Also,

$$C=\{T\subseteq A:\text{ sum of elements of }T\text{ is prime}\}$$

We need

$$n(B\cup C)$$

First find

$$n(B)$$

Total subsets of $$A$$ are

$$2^7=128$$

Now complement of $$B$$ occurs when

$$1\in T\quad \text{and}\quad 2\notin T$$

The remaining elements $$3,4,5,6,7$$ can be chosen freely.

Hence,

$$n(B')=2^5=32$$

Therefore,

$$n(B)=128-32=96$$

Now find subsets in $$B'$$ whose sum is prime.

For sets in $$B',$$

$$1\in T,\qquad 2\notin T$$

So write

$$T=\{1\}\cup S,$$

where

$$S\subseteq\{3,4,5,6,7\}$$

Now total sum becomes

$$1+\text{sum}(S)$$

Possible prime totals are

$$5,7,11,13,17,19,23,29$$

Hence required subset sums are

$$4,6,10,12,16,18,22,28$$

Now count them.

Sum $$4:$$

$$\{4\}\quad\Rightarrow1$$ case

Sum $$6:$$

$$\{6\}\quad\Rightarrow1$$ case

Sum $$10:$$

$$\{3,7\},\{4,6\}\quad\Rightarrow2$$ cases

Sum $$12:$$

$$\{5,7\}\quad\Rightarrow1$$ case

Sum $$16:$$

$$\{3,6,7\},\{4,5,7\}\quad\Rightarrow2$$ cases

Sum $$18:$$

$$\{5,6,7\}\quad\Rightarrow1$$ case

Sum $$22:$$

$$\{3,4,5,7\},\{4,5,6,7\}\quad\Rightarrow2$$ cases

Sum $$28:$$

$$\{3,4,5,6,7\}\quad\Rightarrow1$$ case

Therefore,

$$n(C\cap B')=1+1+2+1+2+1+2+1$$

$$=11$$

Hence,

$$n(B\cup C)=128-(32-11)$$

$$=128-21$$

$$=107$$

Therefore, the required number of elements is

$$\boxed{107}$$.

We need to find the total number of 3-digit numbers whose GCD with 36 is 2.

First, factorizing 36 gives $$36 = 2^2 \times 3^2$$.

Next, for $$\gcd(n, 36) = 2$$, we require $$n$$ to be even but not divisible by 4 (so that the highest power of 2 dividing the GCD is exactly $$2^1$$), and also not divisible by 3. In other words, $$n = 2m$$ where $$m$$ is odd and not divisible by 3.

Now, considering 3-digit numbers with $$100 \le 2m \le 999$$, we get $$50 \le m \le 499$$.

Since $$m$$ must be odd and not divisible by 3, we first count the odd integers from 51 to 499, which number $$\frac{499 - 51}{2} + 1 = 225$$.

Subsequently, we remove the odd multiples of 3 in the same range. These are numbers of the form $$3k$$ where $$k$$ is odd and $$51 \le 3k \le 499$$. The smallest such number is 51 ($$= 3 \times 17$$) and the largest is 495 ($$= 3 \times 165$$), giving the sequence $$3 \times 17, 3 \times 19, \dots, 3 \times 165$$, which has $$\frac{165 - 17}{2} + 1 = 75$$ terms.

Therefore, the total count is $$225 - 75 = 150$$, so the answer is $$150$$.

Given,

$$2f(a)-f(b)+3f(c)+f(d)=0$$

Rearranging,

$$f(b)=2f(a)+3f(c)+f(d)$$

Since

$$f:\{a,b,c,d\}\to\{0,1,2,\ldots,10\}$$

is one-one, all values

$$f(a),f(b),f(c),f(d)$$

must be distinct.

Also,

$$f(b)\le10$$

Hence,

$$2f(a)+3f(c)+f(d)\le10$$

Now consider cases according to $$f(c).$$

Case 1: $$f(c)=0$$

Then,

$$f(b)=2f(a)+f(d)$$

Since values are distinct,

$$f(a),f(d)\neq0$$

Now count possible ordered pairs $$\big(f(a),f(d)\big).$$

For $$f(a)=1,$$

$$2+f(d)\le10\implies f(d)\le8$$

Excluding $$f(d)=1,$$

number of choices $$=7$$

For $$f(a)=2,$$

$$4+f(d)\le10\implies f(d)\le6$$

Excluding $$f(d)=2,$$

number of choices $$=5$$

For $$f(a)=3,$$

$$6+f(d)\le10\implies f(d)\le4$$

Excluding $$f(d)=3,$$

number of choices $$=3$$

For $$f(a)=4,$$

$$8+f(d)\le10\implies f(d)\le2$$

Number of choices $$=2$$

Total cases,

$$7+5+3+2=17$$

Case 2: $$f(c)=1$$

Then,

$$f(b)=2f(a)+f(d)+3$$

Now values cannot equal $$1.$$

For $$f(a)=0,$$

$$f(d)+3\le10\implies f(d)\le7$$

Excluding $$0,1,$$

choices $$=6$$

For $$f(a)=2,$$

$$4+f(d)+3\le10\implies f(d)\le3$$

Excluding $$1,2,$$

choices $$=2$$

For $$f(a)=3,$$

$$6+f(d)+3\le10\implies f(d)\le1$$

Excluding $$1,$$

choices $$=1$$

Total cases,

$$6+2+1=9$$

Case 3: $$f(c)=2$$

Then,

$$f(b)=2f(a)+f(d)+6$$

For $$f(a)=0,$$

$$f(d)+6\le10\implies f(d)\le4$$

Excluding $$0,2,$$

choices $$=3$$

For $$f(a)=1,$$

$$2+f(d)+6\le10\implies f(d)\le2$$

Excluding $$1,2,$$

choices $$=1$$

Total cases,

$$3+1=4$$

Case 4: $$f(c)=3$$

Then,

$$f(b)=2f(a)+f(d)+9$$

For $$f(a)=0,$$

$$f(d)+9\le10\implies f(d)\le1$$

Excluding $$0,$$

choices $$=1$$

Total cases $$=1$$

Therefore, total number of one-one functions is

$$17+9+4+1=31$$

Hence, the required number is

$$\boxed{31}$$.

We need to count $$2 \times 2$$ matrices with entries from $$\{0, 1, 2, 3, 4, 5\}$$ whose entry sum is a prime $$p$$ with $$2 < p < 8$$.

The primes in the range $$(2, 8)$$ are $$p = 3, 5, 7$$.

We need the number of solutions to $$a + b + c + d = p$$ where $$0 \leq a, b, c, d \leq 5$$.

Case 1: $$p = 3$$

Without the upper bound constraint, the number of non-negative integer solutions is $$\binom{3+3}{3} = \binom{6}{3} = 20$$.

Since no variable can exceed 3 when the sum is 3, all solutions satisfy the upper bound. Count = 20.

Case 2: $$p = 5$$

Unrestricted: $$\binom{8}{3} = 56$$.

Since the sum is 5, no single variable can exceed 5, so the upper bound constraint is automatically satisfied. Count = 56.

Case 3: $$p = 7$$

Unrestricted: $$\binom{10}{3} = 120$$.

Subtract cases where any variable $$\geq 6$$: if $$a \geq 6$$, let $$a' = a - 6$$, then $$a' + b + c + d = 1$$, giving $$\binom{4}{3} = 4$$ solutions.

By symmetry across all 4 variables: subtract $$4 \times 4 = 16$$.

Count = $$120 - 16 = 104$$.

Total:

$$20 + 56 + 104 = 180$$

The correct answer is $$\boxed{180}$$.

We need to count $$3 \times 3$$ matrices with entries from $$\{-1, 0, 1\}$$ whose entries sum to $$5$$.

Let $$k$$ be the number of entries equal to 1, $$m$$ be the number equal to $$-1$$, and $$9 - k - m$$ the number equal to 0. We require $$k - m = 5$$ and $$k + m \le 9$$ with $$k, m \ge 0$$. Since $$k = m + 5$$, substituting yields $$2m + 5 \le 9$$, hence $$m \le 2$$.

When $$m = 0$$, it follows that $$k = 5$$ and there are 4 zeros. Choosing 5 positions for 1’s among 9 gives $$\binom{9}{5} = 126$$.

For $$m = 1$$, we have $$k = 6$$ and 2 zeros; thus we choose 6 positions for 1’s and 1 of the remaining 3 positions for a $$-1$$, giving $$\binom{9}{6} \times \binom{3}{1} = 84 \times 3 = 252$$.

When $$m = 2$$, then $$k = 7$$ and there are no zeros, so we select 7 positions for 1’s (the remaining 2 are $$-1$$’s), yielding $$\binom{9}{7} = 36$$.

Adding these counts gives $$126 + 252 + 36 = 414$$, and therefore the total number of such matrices is $$414$$.

We need to count $$3 \times 3$$ matrices with entries 0 or 1 such that the sum of all 9 entries is a prime number. Each entry is independently 0 or 1, so the sum of entries ranges from 0 to 9.

The prime values in $$\{0, 1, 2, \ldots, 9\}$$ are $$\{2, 3, 5, 7\}$$.

The number of matrices with entry sum equal to $$k$$ is $$\binom{9}{k}$$ (choosing which $$k$$ of the 9 entries are 1). So the total count is:

$$\binom{9}{2} + \binom{9}{3} + \binom{9}{5} + \binom{9}{7}$$

We compute each: $$\binom{9}{2} = 36$$, $$\binom{9}{3} = 84$$, $$\binom{9}{5} = 126$$, $$\binom{9}{7} = 36$$.

The total is $$36 + 84 + 126 + 36 = 282$$.

Hence, the correct answer is 282.

We need to find the sum $${}^{n+1}C_2 + 2\left({}^2C_2 + {}^3C_2 + {}^4C_2 + \ldots + {}^nC_2\right)$$.

First, we compute $${}^{n+1}C_2 = \frac{n(n+1)}{2}$$.

For the inner sum, we apply the hockey stick identity: $${}^2C_2 + {}^3C_2 + \ldots + {}^nC_2 = {}^{n+1}C_3 = \frac{(n+1)n(n-1)}{6}$$.

Substituting into the original expression, we get $$\frac{n(n+1)}{2} + 2 \cdot \frac{(n+1)n(n-1)}{6} = \frac{n(n+1)}{2} + \frac{n(n+1)(n-1)}{3}$$.

Taking $$n(n+1)$$ as a common factor, this becomes $$n(n+1)\left(\frac{1}{2} + \frac{n-1}{3}\right) = n(n+1) \cdot \frac{3 + 2(n-1)}{6} = \frac{n(n+1)(2n+1)}{6}$$.

Therefore, the sum equals $$\frac{n(n+1)(2n+1)}{6}$$.

Set $$A$$ has 3 elements and set $$B$$ has 5 elements. The number of one-one functions from $$A$$ to $$B$$ is $$x = P(5, 3) = 5 \times 4 \times 3 = 60$$.

The set $$A \times B$$ has $$3 \times 5 = 15$$ elements. The number of one-one functions from $$A$$ (3 elements) to $$A \times B$$ (15 elements) is $$y = P(15, 3) = 15 \times 14 \times 13 = 2730$$.

Now we check the ratio: $$\frac{2y}{x} = \frac{2 \times 2730}{60} = \frac{5460}{60} = 91$$.

Therefore, $$2y = 91x$$.

We begin by writing the six letters of the word FARMER in strict alphabetical order. In English dictionary order we have $$A \lt E \lt F \lt M \lt R.$$ Remember that the letter $$R$$ occurs twice while every other letter occurs once.

In the required list we include all six-letter arrangements of these letters except those in which the two $$R$$’s stand next to each other. Our task is to find how many valid words appear before the given word FARMER, and then add one for FARMER itself.

Step 1 : fixing the first letter. The first letter of FARMER is $$F$$. All words that start with letters alphabetically smaller than $$F$$ will certainly come before FARMER. The letters smaller than $$F$$ are $$A$$ and $$E$$ (two letters).

Suppose we fix such a smaller letter (say $$A$$) in the first place. We must then arrange the remaining five letters $$E,\,F,\,M,\,R,\,R$$ without letting the two $$R$$’s touch.

We first arrange the three non-$$R$$ letters $$E,\,F,\,M$$ in every possible order. Because they are all distinct, the number of such orders is $$3! = 6.$$

Whenever these three letters are written, they create $$4$$ “gaps’’ (one before the first letter, two between consecutive letters, and one after the last letter). To keep the two $$R$$’s apart, we simply choose any two of these four gaps and drop one $$R$$ into each chosen gap. Because the $$R$$’s are identical, the only decision is the choice of gaps, which can be done in $$\binom{4}{2} = 6$$ ways.

Hence the total number of valid words that start with a fixed smaller letter such as $$A$$ is $$3! \times \binom{4}{2} \;=\; 6 \times 6 \;=\; 36.$$

The same calculation holds if the first letter is $$E$$, so we obtain

$$36 + 36 = 72$$

valid words whose first letter precedes $$F$$. All those 72 words will appear before FARMER.

Step 2 : fixing the first two letters as $$F A$$. With the first letter now forced to be $$F$$, we turn to the second position. The second letter in FARMER is $$A$$. Among the unused letters $$A,\,E,\,M,\,R,\,R$$ none is alphabetically smaller than $$A$$, so no additional words are formed here. Our running count stays at $$72$$ words.

Step 3 : fixing the first three letters as $$F A$$ and choosing the third. The third letter in FARMER is $$R$$. The unused letters are $$E,\,M,\,R,\,R$$. Letters alphabetically smaller than $$R$$ are $$E$$ and $$M$$. We look at each choice in turn.

(i) Take $$E$$ as the third letter, giving the prefix $$F A E$$. The remaining letters are $$M,\,R,\,R$$. Among the $$3!/2! = 3$$ ordinary permutations of these three letters, only $$R M R$$ keeps the two $$R$$’s apart, so exactly one valid word is produced.

(ii) Take $$M$$ as the third letter, yielding the prefix $$F A M$$. The remaining letters are $$E,\,R,\,R$$, and again only $$R E R$$ is acceptable. Thus a second valid word is obtained.

Hence

$$1 + 1 = 2$$

additional words appear before any word that actually has $$F A R$$ as its first three letters. Adding these two words raises the count to

$$72 + 2 = 74.$$

Step 4 : fixing the prefix $$F A R$$ and choosing the fourth letter. The fourth letter in FARMER is $$M$$. After the prefix $$F A R$$ the unused letters are $$E,\,M,\,R$$. Among these, the letter alphabetically smaller than $$M$$ is only $$E$$.

If we select $$E$$ for the fourth place, the prefix becomes $$F A R E$$ and the leftover letters are $$M$$ and $$R$$. Because there is now only one $$R$$ remaining, there is no danger of placing two $$R$$’s together. The two distinct letters $$M$$ and $$R$$ can be arranged in

$$2! = 2$$

ways. Thus two more valid words precede FARMER, bringing the running total to

$$74 + 2 = 76.$$

Step 5 : fixing the prefix $$F A R M$$. With the fourth letter matching FARMER, the remaining letters are $$E$$ and $$R$$. The fifth letter in FARMER is $$E$$. Since no unused letter is alphabetically smaller than $$E$$, no new words are gained here.

The prefix now exactly matches $$F A R M E$$, leaving one final $$R$$ to be placed, which is allowed because it sits after an $$E$$ and is therefore not adjacent to another $$R$$.

Step 6 : counting FARMER itself. All the words counted so far (76 of them) precede FARMER. Consequently, the word FARMER occupies position

$$76 + 1 = 77.$$

So, the answer is $$77$$.

We decompose the expression $$r^3 + 6r^2 + 2r + 5$$. Since $$(r+1)(r+2)(r+3) = r^3 + 6r^2 + 11r + 6$$, we have $$r^3 + 6r^2 + 2r + 5 = (r+1)(r+2)(r+3) - 9r - 1$$.

Multiplying by $$r!$$ gives $$r!(r^3 + 6r^2 + 2r + 5) = r!(r+1)(r+2)(r+3) - 9r \cdot r! - r!$$. Now $$r!(r+1)(r+2)(r+3) = (r+3)!$$, and using the identity $$r \cdot r! = (r+1)! - r!$$, we get $$(r+3)! - 9[(r+1)! - r!] - r! = (r+3)! - 9(r+1)! + 8 \cdot r!$$.

Each of the three sums telescopes when summed from $$r = 1$$ to $$10$$. For the first: $$\sum_{r=1}^{10}(r+3)! = 4! + 5! + \cdots + 13!$$. For the second: $$9\sum_{r=1}^{10}(r+1)! = 9(2! + 3! + \cdots + 11!)$$. For the third: $$8\sum_{r=1}^{10}r! = 8(1! + 2! + \cdots + 10!)$$.

We compute each sum step by step. Writing $$A_n = 1! + 2! + \cdots + n!$$, we have $$A_{10} = 4{,}037{,}913$$. Then: $$\sum_{r=1}^{10}(r+3)! = A_{13} - A_3 = (A_{10} + 11! + 12! + 13!) - (1! + 2! + 3!) = 4{,}037{,}913 + 39{,}916{,}800 + 479{,}001{,}600 + 6{,}227{,}020{,}800 - 9 = 6{,}749{,}977{,}104$$. Similarly, $$9 \cdot (A_{11} - A_1) = 9(4{,}037{,}913 + 39{,}916{,}800 - 1) = 9 \times 43{,}954{,}712 = 395{,}592{,}408$$. And $$8 \cdot A_{10} = 8 \times 4{,}037{,}913 = 32{,}303{,}304$$.

Therefore the total sum is $$6{,}749{,}977{,}104 - 395{,}592{,}408 + 32{,}303{,}304 = 6{,}386{,}688{,}000$$. Since $$11! = 39{,}916{,}800$$, we get $$\alpha = \dfrac{6{,}386{,}688{,}000}{39{,}916{,}800} = 160$$.

We have the set $$S=\{1,2,3,4,5,6,9\}$$ containing seven elements. For every subset $$A\subseteq S$$ we are interested in the remainder obtained when the sum of its elements is divided by 3. Because divisibility by 3 depends only on the remainder, it is natural to classify each element of $$S$$ according to its remainder (or “residue”) modulo 3.

Computing the residues, we obtain

$$\begin{aligned} 1&\equiv1\pmod3,\\ 2&\equiv2\pmod3,\\ 3&\equiv0\pmod3,\\ 4&\equiv1\pmod3,\\ 5&\equiv2\pmod3,\\ 6&\equiv0\pmod3,\\ 9&\equiv0\pmod3. \end{aligned}$$

So the seven elements fall into three residue classes:

$$ \begin{array}{lcl} \text{Class }0:&\; \{3,6,9\} &\Longrightarrow n_0=3,\\[2pt] \text{Class }1:&\; \{1,4\} &\Longrightarrow n_1=2,\\[2pt] \text{Class }2:&\; \{2,5\} &\Longrightarrow n_2=2. \end{array} $$

Choosing elements from the “0‐class” never alters the remainder, because each of those numbers is itself a multiple of 3. Hence any subset chosen from $$\{3,6,9\}$$ contributes $$0$$ (mod 3) to the total sum. There are $$2^{n_0}=2^{3}=8$$ possible selections from this class, including the empty selection.

If we pick $$k_1$$ elements from the “1‐class”, each contributes a remainder $$1$$, so their combined remainder is

$$k_1\pmod3.$$

Because $$n_1=2$$, the value $$k_1$$ may be $$0,1,$$ or $$2$$, and the corresponding numbers of ways are

$$\binom{2}{0}=1,\quad\binom{2}{1}=2,\quad\binom{2}{2}=1.$$

Similarly, choosing $$k_2$$ elements from the “2‐class” gives a combined remainder

$$2k_2\pmod3,$$

and with $$n_2=2$$ the admissible values and counts are also

$$\binom{2}{0}=1,\quad\binom{2}{1}=2,\quad\binom{2}{2}=1.$$

For the total sum of a subset to be a multiple of 3 we require

$$k_1+2k_2\equiv0\pmod3.$$

We now list every pair $$(k_1,k_2)$$ satisfying this condition together with the number of ways it can occur.

1. $$k_1=0$$ gives residue $$0$$. We need $$2k_2\equiv0\pmod3$$, which is true only for $$k_2=0$$. Number of ways:

$$\binom{2}{0}\times\binom{2}{0}=1\times1=1.$$

2. $$k_1=1$$ gives residue $$1$$. We need $$1+2k_2\equiv0\pmod3\;\Longrightarrow\;2k_2\equiv2\pmod3\;\Longrightarrow\;k_2=1.$$ Number of ways:

$$\binom{2}{1}\times\binom{2}{1}=2\times2=4.$$

3. $$k_1=2$$ gives residue $$2$$. We need $$2+2k_2\equiv0\pmod3\;\Longrightarrow\;2k_2\equiv1\pmod3\;\Longrightarrow\;k_2=2.$$ Number of ways:

$$\binom{2}{2}\times\binom{2}{2}=1\times1=1.$$

Adding these counts, the total number of ways to choose elements from the “1‐class” and “2‐class” so that the running sum is divisible by 3 is

$$1+4+1=6.$$

For every such choice we may freely choose any subset of the three “0‐class” elements, and there are $$8$$ possibilities for that. Multiplying, the number of all subsets (including the empty one) whose element-sum is a multiple of 3 equals

$$6\times8=48.$$

Among these $$48$$ subsets, one is the empty set itself. Because the problem specifically excludes the empty set (it demands $$A\neq\varnothing$$), the count of non-empty subsets with sum divisible by 3 is

$$48-1=47.$$

The total number of non-empty subsets of a 7-element set is

$$2^{7}-1=128-1=127.$$

Therefore the number of non-empty subsets whose element-sum is not a multiple of 3 is obtained by simple subtraction:

$$127-47=80.$$

Hence, the correct answer is Option C ($$80$$).

We need the sum of all $$n \in \{1, 2, \ldots, 100\}$$ such that $$\gcd(n, 2040) = 1$$. First, factorize $$2040 = 2^3 \times 3 \times 5 \times 17$$. So $$n$$ must not be divisible by 2, 3, 5, or 17.

We use inclusion-exclusion on the sum. Let $$S_d$$ denote the sum of all multiples of $$d$$ in $$\{1, \ldots, 100\}$$. If there are $$k = \lfloor 100/d \rfloor$$ such multiples, then $$S_d = d \cdot \frac{k(k+1)}{2}$$.

The sum we want is $$S_{\text{total}} - (S_2 + S_3 + S_5 + S_{17}) + (S_6 + S_{10} + S_{34} + S_{15} + S_{51} + S_{85}) - (S_{30} + S_{102} + S_{170} + S_{255}) + S_{510}$$, where terms with $$d > 100$$ contribute zero.

Computing each: the total sum is $$\frac{100 \times 101}{2} = 5050$$. For $$S_2$$: $$k = 50$$, $$S_2 = 2 \times 1275 = 2550$$. For $$S_3$$: $$k = 33$$, $$S_3 = 3 \times 561 = 1683$$. For $$S_5$$: $$k = 20$$, $$S_5 = 5 \times 210 = 1050$$. For $$S_{17}$$: $$k = 5$$, $$S_{17} = 17 \times 15 = 255$$.

For the pairwise terms: $$S_6$$: $$k = 16$$, $$S_6 = 6 \times 136 = 816$$. $$S_{10}$$: $$k = 10$$, $$S_{10} = 10 \times 55 = 550$$. $$S_{34}$$: $$k = 2$$, $$S_{34} = 34 \times 3 = 102$$. $$S_{15}$$: $$k = 6$$, $$S_{15} = 15 \times 21 = 315$$. $$S_{51}$$: $$k = 1$$, $$S_{51} = 51$$. $$S_{85}$$: $$k = 1$$, $$S_{85} = 85$$.

For the triple terms: $$S_{30}$$: $$k = 3$$, $$S_{30} = 30 \times 6 = 180$$. $$S_{102}$$, $$S_{170}$$, $$S_{255}$$ all have $$k = 0$$, contributing nothing. The quadruple term $$S_{510} = 0$$ as well.

Putting it all together: $$5050 - (2550 + 1683 + 1050 + 255) + (816 + 550 + 102 + 315 + 51 + 85) - 180 = 5050 - 5538 + 1919 - 180 = 1251$$.

We have to count those positive divisors that can be written in the form $$4n+1$$ for the number $$ (10)^{10}\,(11)^{11}\,(13)^{13}. $$ First we write every factor in terms of its prime constituents:

$$10 = 2\cdot5,$$ so $$ (10)^{10}=2^{10}\,5^{10}.$$ Hence the complete prime‐factorisation of the given number is

$$ N \;=\;2^{10}\,5^{10}\,11^{11}\,13^{13}. $$

Any divisor $$d$$ of $$N$$ therefore has the general form

$$ d \;=\;2^{a}\,5^{b}\,11^{c}\,13^{d_1}, $$

where the exponents are restricted by the powers present in $$N$$:

$$ 0\le a\le 10,\quad 0\le b\le 10,\quad 0\le c\le 11,\quad 0\le d_1\le 13. $$

To ensure that $$d$$ is of the form $$4n+1$$ we must have $$d\equiv1\pmod4.$$ We therefore study each prime factor modulo $$4$$:

$$ 2\equiv2\pmod4,\qquad5\equiv1\pmod4,\qquad11\equiv3\pmod4,\qquad13\equiv1\pmod4. $$

Because $$5\equiv1\pmod4$$ and $$13\equiv1\pmod4,$$ every power of $$5$$ or $$13$$ is again congruent to $$1$$ modulo $$4$$:

$$ 5^{b}\equiv1\pmod4,\qquad13^{d_1}\equiv1\pmod4\quad\text{for all }b,d_1. $$

Only the powers of $$2$$ and $$11$$ can change the final remainder. We therefore examine them separately.

Behaviour of $$2^{a}\pmod4$$

$$ 2^{0}=1,\quad2^{1}=2,\quad2^{2}=4\equiv0\pmod4, $$ and for any $$a\ge2$$ we still have $$2^{a}\equiv0\pmod4.$$ Hence

$$ 2^{a}\equiv \begin{cases} 1,&a=0,\\ 2,&a=1,\\ 0,&a\ge2. \end{cases} $$

If $$a\ge2$$ the factor $$2^{a}$$ already forces the whole product to be $$0\pmod4,$$ so such divisors can never be congruent to $$1\pmod4.$$ Therefore we must restrict ourselves to

$$ a=0\quad\text{or}\quad a=1. $$

Behaviour of $$11^{c}\pmod4$$

Since $$11\equiv3\pmod4,$$ we use the basic property $$3^{2}\equiv9\equiv1\pmod4.$$ Thus

$$ 11^{c}\equiv3^{c}\equiv \begin{cases} 1,&c\text{ even},\\ 3,&c\text{ odd}. \end{cases} $$

Combining the two factors

The total remainder contributed by $$2^{a}11^{c}$$ is

$$ (2^{a}\bmod4)\times(3^{c}\bmod4)\pmod4. $$

We now test the only possible values of $$a$$(namely $$0$$ and $$1$$):

• If $$a=0,$$ then $$2^{a}\equiv1,$$ so we need $$ 1\times3^{c}\equiv1\pmod4, $$ which is true exactly when $$3^{c}\equiv1,$$ i.e. when $$c$$ is even.

• If $$a=1,$$ then $$2^{a}\equiv2,$$ giving $$ 2\times3^{c}\pmod4\in\{2,\,2\}\quad(\text{because }3^{c}\equiv1\text{ or }3), $$ both of which are $$2\pmod4,$$ never $$1.$$ Hence no divisor with $$a=1$$ can satisfy the condition.

Consequently the required divisors must have

$$ a=0,\qquad c\text{ even}. $$

Counting the allowable choices

• For $$a$$ there is only one choice: $$a=0.$$
• For $$c$$ (ranging from $$0$$ to $$11$$) the even values are $$0,2,4,6,8,10,$$ a total of $$6$$ choices.
• For $$b$$ any value from $$0$$ to $$10$$ is permitted, giving $$10-0+1=11$$ choices.
• For $$d_1$$ any value from $$0$$ to $$13$$ is allowed, giving $$13-0+1=14$$ choices.

Multiplying the independent counts we obtain the total number of divisors congruent to $$1\pmod4$$:

$$ 1\times6\times11\times14 \;=\; 6\times154 \;=\; 924. $$

So, the answer is $$924$$.

We have to evaluate the expression

$$1\cdot{}^1P_1+2\cdot{}^2P_2+3\cdot{}^3P_3+\ldots+15\cdot{}^{15}P_{15}$$

and compare it with $$^qP_r-s$$ where $$0\le s\le1$$. After finding $$q,\;r,\;s$$ we shall compute $$^{\,q+s}C_{\,r-s}$$.

First recall the definition of permutations:

$$^nP_r=\frac{n!}{(n-r)!}.$$

In every term of our sum the two numbers are the same, i.e. $$r=n$$, so we get

$$^kP_k=\frac{k!}{(k-k)!}=\frac{k!}{0!}=k!.$$

Hence every term simplifies to $$k\cdot k!$$ and the entire left‐hand side becomes

$$\sum_{k=1}^{15}k\cdot k!.$$

There is a standard identity:

$$\sum_{k=1}^{n}k\cdot k!=(n+1)!-1.$$ This can be proved quickly by mathematical induction, but we shall accept it as known.

Putting $$n=15$$ we obtain

$$\sum_{k=1}^{15}k\cdot k!=(15+1)!-1=16!-1.$$

So the given equality becomes

$$^qP_r-s=16!-1.$$

Now $$s$$ can be only $$0$$ or $$1$$. If $$s=0$$ we would need $$^qP_r=16!-0=16!-1,$$ which is not a factorial and very unlikely to be a permutation value. Choosing $$s=1$$ gives

$$^qP_r=16!.$$

To realise a value of exactly $$16!$$ with a permutation, the simplest way is to take $$r=q$$. Then

$$^qP_q=q!,$$

and the equation $$q!=16!$$ forces

$$q=16,\qquad r=16.$$

Therefore we have found

$$q=16,\qquad r=16,\qquad s=1.$$

We are asked to compute

$$^{\,q+s}C_{\,r-s}.$$

Substituting the values just obtained, we get

$$^{\,q+s}C_{\,r-s}=^{\,16+1}C_{\,16-1}=^{17}C_{15}.$$

The binomial coefficient satisfies $$^nC_r=^nC_{n-r},$$ so

$$^{17}C_{15}=^{17}C_{2}.$$

Using the combination formula

$$^nC_r=\frac{n!}{r!\,(n-r)!},$$

we calculate

$$^{17}C_2=\frac{17!}{2!\,15!}=\frac{17\cdot16\cdot15!}{2\cdot15!}=\frac{17\cdot16}{2}=136.$$

Hence, the correct answer is Option 136.

We need bijective functions $$f : A \to A$$ where $$A = \{0, 1, 2, 3, 4, 5, 6, 7\}$$ satisfying $$f(1) + f(2) = 3 - f(3)$$, i.e., $$f(1) + f(2) + f(3) = 3$$.

Since $$f$$ is a bijection (permutation of $$A$$), the values $$f(1), f(2), f(3)$$ must be three distinct elements of $$A$$ that sum to 3.

We need to find all sets of three distinct non-negative integers from $$\{0, 1, 2, 3, 4, 5, 6, 7\}$$ that sum to 3. The possible unordered sets are:

$$\{0, 1, 2\}$$: sum $$= 0 + 1 + 2 = 3$$. This works.

No other triple of distinct non-negative integers from $$A$$ sums to 3, since the next smallest sum of three distinct elements would be $$0 + 1 + 3 = 4$$.

So $$\{f(1), f(2), f(3)\} = \{0, 1, 2\}$$. The three values can be assigned to $$f(1), f(2), f(3)$$ in $$3! = 6$$ ways.

The remaining 5 inputs $$\{0, 4, 5, 6, 7\}$$ must map to the remaining 5 outputs $$\{3, 4, 5, 6, 7\}$$, and these can be assigned in $$5! = 120$$ ways.

The total number of bijective functions is $$6 \times 120 = 720$$.

We have to count the five-digit natural numbers in which every digit is different and the digit at the tens place (that is, the $$10^{1}$$ place) is a fixed 2.

Let us write a general five-digit number as

$$\overline{d_5\,d_4\,d_3\,d_2\,d_1},$$

where $$d_5$$ is the ten-thousands digit, $$d_4$$ the thousands digit, $$d_3$$ the hundreds digit, $$d_2$$ the tens digit and $$d_1$$ the units digit. According to the statement $$d_2=2$$. All the remaining digits have to be chosen from the other ten numerals $$0,1,3,4,5,6,7,8,9$$ with the restriction that no digit repeats.

Step 1. Choosing $$d_5$$. For a five-digit number the leading digit cannot be $$0$$, and it also cannot be $$2$$ (because $$2$$ is already sitting at the tens place). Hence $$d_5$$ can be chosen from the set $$\{1,3,4,5,6,7,8,9\}$$, which gives

$$8\text{ ways.}$$

Step 2. Choosing $$d_4$$. After we have fixed $$d_5$$ and $$d_2$$, two different digits are already used. Eight digits are still unused: $$0$$ together with the seven digits that are not $$2$$ and not equal to the chosen $$d_5$$. Therefore

$$d_4$$ can be filled in $$8\text{ ways.}$$

Step 3. Choosing $$d_3$$. Now three different digits have been occupied, so

$$d_3$$ can be selected in $$7\text{ ways.}$$

Step 4. Choosing $$d_1$$. With four distinct digits already fixed, six digits remain unused, and any of them may be taken for the units place. Hence

$$d_1$$ can be assigned in $$6\text{ ways.}$$

Total count. By the multiplication principle, the total number of admissible five-digit numbers is

$$8 \times 8 \times 7 \times 6 \;=\;2688.$$

This number is expressed in the statement as $$336k$$, so

$$336k \;=\;2688 \;\;\Longrightarrow\;\; k \;=\;\dfrac{2688}{336} \;=\;8.$$

Hence, the correct answer is Option D.

We have $$n$$ Metro stations placed on a circle, with $$n > 2$$. For any two stations we draw exactly one straight track, so overall the network is the complete graph on $$n$$ vertices.

Among these tracks, those joining the two stations that are next to each other on the circle are declared blue. Because the stations form a single closed polygon, every station is adjacent to exactly two neighbours, but every such edge is counted only once. Hence the number of blue tracks equals the number of sides of the polygon, namely

$$\text{Blue lines}=n.$$

All the other tracks, i.e. those whose ends are not nearest neighbours on the circle, are coloured red. Let us count how many tracks there are altogether. In a complete graph on $$n$$ vertices, the number of unordered pairs of vertices (and hence the number of tracks) is given by the combination formula

$$\binom{n}{2}= \frac{n(n-1)}{2}.$$

Therefore

$$\text{Total tracks}= \frac{n(n-1)}{2}.$$ $$\text{Red lines}= \text{Total tracks}-\text{Blue lines}= \frac{n(n-1)}{2}-n.$$

The condition given in the question says that the number of red lines is 99 times the number of blue lines, that is,

$$\frac{n(n-1)}{2}-n = 99\,n.$$

Now we solve this equation step by step. First clear the fraction by multiplying both sides by 2:

$$n(n-1) - 2n = 198\,n.$$

Expand the left side:

$$n^2 - n - 2n = 198\,n.$$

Combine like terms on the left:

$$n^2 - 3n = 198\,n.$$

Bring the right‐hand side to the left:

$$n^2 - 3n - 198\,n = 0,$$ $$n^2 - 201\,n = 0.$$

Factor out $$n$$:

$$n\,(n-201)=0.$$

Since we are told $$n>2$$, the factor $$n=0$$ is impossible. Hence we must have

$$n-201 = 0 \quad\Longrightarrow\quad n = 201.$$

Hence, the correct answer is Option A.

Let the three sections be labelled Section I, Section II and Section III. If we denote by $$x_1,\,x_2,\,x_3$$ the number of questions chosen from Section I, Section II and Section III respectively, then

$$x_1+x_2+x_3 = 5,$$

because a total of five questions have to be answered, and

$$x_1 \ge 1,\; x_2 \ge 1,\; x_3 \ge 1,$$

because the candidate must attempt at least one question from each section. We now list all integer triples $$\,(x_1,x_2,x_3)\,$$ that satisfy these two conditions.

First we rewrite the condition $$x_1+x_2+x_3 = 5$$ under the restriction $$x_i \ge 1$$. If we set $$y_i = x_i-1$$ for $$i=1,2,3$$, then $$y_i \ge 0$$ and

$$y_1 + y_2 + y_3 = 5 - 3 = 2.$$

Hence we need the non-negative solutions of $$y_1 + y_2 + y_3 = 2$$. Using the stars-and-bars formula, the total number of such solutions is

$$\binom{2+3-1}{3-1} = \binom{4}{2} = 6,$$

and they are explicitly

$$$(2,0,0),\, (0,2,0),\, (0,0,2),\, (1,1,0)$$$, $$(1,0,1),\, (0,1,1).$$

Translating back by adding 1 to every component, the admissible triples $$\,(x_1,x_2,x_3)\,$$ are

$$$(3,1,1),\, (1,3,1),\, (1,1,3),\, (2,2,1)$$$, $$(2,1,2),\, (1,2,2).$$

So there are exactly two distinct types of distributions:

Type A. One section contributes 3 questions and each of the other two contributes 1 question  →  patterns $$\;(3,1,1)\,$$, $$\,(1,3,1)\,$$, $$\,(1,1,3).$$

Type B. Two sections contribute 2 questions each and the remaining one contributes 1 question  →  patterns $$\;(2,2,1)\,$$, $$\,(2,1,2)\,$$, $$\,(1,2,2).$$

We now count the actual choices of questions for every pattern.

Type A: pattern (3,1,1). From the section where 3 questions are to be chosen, the number of ways is $$\binom{5}{3}.$$ From each section where 1 question is to be chosen, the number of ways is $$\binom{5}{1}.$$ Hence, for this pattern, the number of ways is

$$$\binom{5}{3}\,\binom{5}{1}\,\binom{5}{1} = 10 \times 5 \times 5 = 250.$$$

The same count applies to each of the 3 permutations of (3,1,1), so the total for Type A is

$$3 \times 250 = 750.$$

Type B: pattern (2,2,1). Here the number of ways is

$$$\binom{5}{2}\,\binom{5}{2}\,\binom{5}{1} = 10 \times 10 \times 5 = 500.$$$

Again there are 3 permutations of (2,2,1), so the total for Type B is

$$3 \times 500 = 1500.$$

Adding the two types together, the grand total number of permissible question selections is

$$750 + 1500 = 2250.$$

Hence, the correct answer is Option D.

We have to count all possible six-digit numbers that are formed with the digits $$1,3,5,7,9$$ under the two simultaneous conditions:

(i) no digit outside this set can occur, and

(ii) each of these five digits must appear at least once in the six positions.

Because we need six places but only five distinct digits, exactly one of the five digits will appear twice and every other digit will appear once.

First we decide which digit is to be repeated. There are $$5$$ choices for this.

After choosing the repeated digit, we decide in which two of the six positions this repeated digit will sit. The number of ways to choose these two positions is given by the combination formula $$ {}^{6}C_{2}=\frac{6!}{2!\,4!}. $$

Once those two places are fixed, the remaining four different digits must fill the remaining four places. Because all four digits are distinct, they can be arranged in $$ 4! $$ ways.

Multiplying the independent choices we get the total required count:

$$ \text{Total} \;=\; 5 \times {}^{6}C_{2} \times 4!. $$

Substituting the values of the factorials, we calculate step by step:

$$ {}^{6}C_{2}=\frac{6!}{2!\,4!}= \frac{720}{2\times24}=15, $$

and

$$ 4!=24. $$

So

$$ \text{Total}=5 \times 15 \times 24. $$

Now multiply sequentially:

$$ 15 \times 24 = 360, $$

and

$$ 5 \times 360 = 1800. $$

For comparison with the given options we rewrite the result using $$6! = 720$$:

$$ 1800 = \frac{5}{2}\times 6!. $$

Hence, the correct answer is Option D.

We have three distinct families: the first family has three members, the second family also has three members, and the third family has four members. The condition “the same family members are not separated” means that every family must sit together as a single block.

So, we first treat each family as one unit or “block.” Now we effectively have only three blocks to arrange in a row.

The number of ways to arrange these three blocks is given by the factorial of the number of blocks. Hence, using the formula for permutations $$n!$$, with $$n = 3$$, we get

$$3!$$

Next, we must consider the internal arrangements within each block.

For the two families that contain three members each, the members can be permuted among themselves independently. The number of ways to arrange the three members of a single family is

$$3!$$

Since there are two such families, the combined number of internal arrangements for these two families is

$$3! \times 3! = (3!)^2$$

Now, for the family that has four members, the members can be arranged among themselves in

$$4!$$

ways.

Because all these internal arrangements happen independently of the arrangement of the blocks, we multiply every individual count together. Hence, the total number of seating arrangements is

$$ \underbrace{3!}_{\text{arrangements of blocks}} \times \underbrace{3!}_{\text{first 3-member family}} \times \underbrace{3!}_{\text{second 3-member family}} \times \underbrace{4!}_{\text{4-member family}} = (3!)^3 \cdot 4! $$

Therefore, the total number of ways is $$(3!)^3 \cdot (4!)$$.

Hence, the correct answer is Option B.

We have to evaluate the following two alternating sums and then add them:

$$S_1 \;=\; 2\cdot {}^1P_0 \;-\; 3\cdot {}^2P_1 \;+\; 4\cdot {}^3P_2 \;-\;\ldots \text{ up to the 51}^{\text{st}}\text{ term},$$

$$S_2 \;=\; 1! \;-\; 2! \;+\; 3! \;-\;\ldots \text{ up to the 51}^{\text{st}}\text{ term}.$$

We first simplify each term of the permutation-based sum $$S_1$$.

Recall the permutation formula:

$$ {}^nP_r \;=\; \dfrac{n!}{(n-r)!}. $$

In the $$k^{\text{th}}$$ term of $$S_1$$ we observe the pattern

$$\text{coefficient} = k+1,\qquad {}^kP_{\,k-1} = \dfrac{k!}{(k-(k-1))!} = \dfrac{k!}{1!} = k!. $$

So the $$k^{\text{th}}$$ term of $$S_1$$ becomes

$$(-1)^{\,k-1}\,(k+1)\,k!,$$

because the sign alternates, starting with $$+$$ for $$k=1$$.

Now we note that $$(k+1)\,k! = (k+1)!$$. Hence

$$S_1 \;=\;\sum_{k=1}^{51} (-1)^{\,k-1}\,(k+1)!.$$

For convenience we shift the index by putting $$j = k+1$$. When $$k$$ runs from $$1$$ to $$51$$, $$j$$ runs from $$2$$ to $$52$$. Thus

$$S_1 \;=\;\sum_{j=2}^{52} (-1)^{\,j-2}\,j!.$$

Next we write the factorial sum $$S_2$$ in sigma notation:

$$S_2 \;=\;\sum_{j=1}^{51} (-1)^{\,j-1}\,j!.$$

The total expression we need is

$$V \;=\; S_1 + S_2 \;=\;\sum_{j=1}^{51} (-1)^{\,j-1}\,j! \;+\;\sum_{j=2}^{52} (-1)^{\,j-2}\,j!.$$

We now combine the two sums term by term.

1. The term with $$j=1$$

Occurs only in the first sum: $$(-1)^{\,1-1}\,1! \;=\; 1\times 1! = 1.$$

2. The terms with $$2 \le j \le 51$$

For every such $$j$$ we have two contributions:

From $$S_2$$: $$(-1)^{\,j-1}\,j!,$$

From $$S_1$$: $$(-1)^{\,j-2}\,j!.$$ Adding them gives $$j!\Bigl[\,(-1)^{\,j-1} + (-1)^{\,j-2}\Bigr] \;=\; j! \,(-1)^{\,j-2}\bigl[\,-1 + 1\bigr] = 0.$$ Thus every factorial from $$2!$$ up to $$51!$$ cancels out completely.

3. The term with $$j=52$$

Appears only in the second sum (shifted $$S_1$$) and equals

$$(-1)^{\,52-2}\,52! \;=\; (-1)^{50}\,52! \;=\; (+1)\,52! = 52!.$$

Collecting the non-zero contributions, we obtain

$$V \;=\; 1 + 52!.$$

Hence, the correct answer is Option C.

We begin with the given equation

$$6\;{}^{35}C_r \;=\;(k^2-3)\;{}^{36}C_{\,r+1}.$$

We wish to connect the two binomial coefficients so that they involve the same upper index. The basic identity for binomial coefficients is

$$^{\,n}C_m \;=\;\dfrac{n}{m}\;{}^{\,n-1}C_{\,m-1}.$$

Using this with $$n=36$$ and $$m=r+1$$ we have

$$^{36}C_{\,r+1} \;=\;\dfrac{36}{r+1}\;{}^{35}C_r.$$

Substituting this into the original equation gives

$$6\;{}^{35}C_r \;=\;(k^2-3)\;\left(\dfrac{36}{r+1}\;{}^{35}C_r\right).$$

Because $$^{35}C_r$$ is non-zero for all integral $$r$$ with $$0\le r\le35,$$ we can cancel it from both sides:

$$6 \;=\;(k^2-3)\;\dfrac{36}{r+1}.$$

Multiplying both sides by $$\dfrac{r+1}{36}$$ yields

$$k^2-3 \;=\;\dfrac{6(r+1)}{36}.$$

Simplifying the fraction $$\dfrac{6}{36}= \dfrac{1}{6},$$ we get

$$k^2-3 \;=\;\dfrac{r+1}{6}.$$

So $$r+1$$ must be a multiple of 6. Write

$$r+1 = 6t,$$

where $$t$$ is an integer. Substituting this into our equation gives

$$k^2-3 = t,$$

or equivalently

$$k^2 = t+3.$$

Next we restrict the possible values of $$t$$. Because $$r$$ must satisfy $$0 \le r \le 35,$$ we have $$1 \le r+1 \le 36,$$ and therefore

$$1 \le 6t \le 36 \;\Longrightarrow\; 1 \le t \le 6.$$

Thus the only permissible values are

$$t = 1,2,3,4,5,6.$$

For each of these $$t$$ we compute $$k^2 = t+3$$:

$$\begin{aligned} t=1 &\;\Rightarrow\; k^2 = 4,\\ t=2 &\;\Rightarrow\; k^2 = 5,\\ t=3 &\;\Rightarrow\; k^2 = 6,\\ t=4 &\;\Rightarrow\; k^2 = 7,\\ t=5 &\;\Rightarrow\; k^2 = 8,\\ t=6 &\;\Rightarrow\; k^2 = 9. \end{aligned}$$

Among the numbers $$4,5,6,7,8,9,$$ the perfect squares are only $$4$$ and $$9.$$ Hence we keep

$$k^2 = 4 \;\text{or}\; k^2 = 9.$$

Handling each case:

• If $$k^2 = 4,$$ then $$k = \pm2.$$ This comes from $$t=1,$$ so $$r+1 = 6\cdot1 = 6 \Rightarrow r = 5.$$ Thus we obtain the two ordered pairs $$(5,2)$$ and $$(5,-2).$$

• If $$k^2 = 9,$$ then $$k = \pm3.$$ This comes from $$t=6,$$ so $$r+1 = 6\cdot6 = 36 \Rightarrow r = 35.$$ Thus we obtain the two ordered pairs $$(35,3)$$ and $$(35,-3).$$

No other $$t$$ values lead to integer $$k$$, so there are altogether

$$2 + 2 = 4$$

ordered pairs $$(r,k).$$

Hence, the correct answer is Option D.

We have six different questions, and each question carries four alternative choices, exactly one of which is right.

First, we decide which questions are answered correctly. Out of the total $$6$$ questions, we want exactly $$4$$ to be correct. The number of ways of choosing these $$4$$ questions is obtained by the combination formula

$$^{n}C_{r}=\dfrac{n!}{r!\,(n-r)!}.$$

Here $$n=6$$ and $$r=4$$, so

$$^{6}C_{4}=\dfrac{6!}{4!\,2!}=15.$$

Now, for each of the chosen $$4$$ questions, there is only one way to answer correctly because only one option is the right one. Thus, the total number of answer patterns for these $$4$$ questions remains $$1^{4}=1$$.

Next, we look at the remaining $$6-4=2$$ questions, which must be answered incorrectly. Each of these questions has $$4-1=3$$ wrong alternatives. Therefore, for every question to be answered wrongly, there are $$3$$ possible choices. Since the two questions are independent, the total number of incorrect answer patterns is

$$3 \times 3 = 3^{2}=9.$$

Finally, we multiply the number of ways of choosing which questions are correct with the number of answer patterns for correct responses and with the number of answer patterns for incorrect responses:

$$\text{Total ways}=^{6}C_{4}\times 1^{4}\times 3^{2}=15\times 1\times 9=135.$$

So, the answer is $$135$$.

First, observe that the word ‘MOTHER’ has six distinct letters. When all six letters are rearranged, we get $$6! = 720$$ different words. These words are to be imagined as written in a dictionary (lexicographic) order, so we must count how many of those words come before the word ‘MOTHER’ itself.

Alphabetically, the six letters are ordered as $$E < H < M < O < R < T.$$ We progress through the word ‘MOTHER’ one position at a time, and at each step we count how many possible words could be formed by choosing a smaller unused letter for that position and permuting the remaining letters arbitrarily.

We denote by “count” the number of such words that appear before ‘MOTHER’.

First position: The first letter of ‘MOTHER’ is $$M.$$ Unused letters smaller than $$M$$ are $$E$$ and $$H,$$ which are two in number. Once one of these two letters is fixed, the remaining five positions can be filled in $$5!$$ ways. Using the factorial rule $$n! = n(n-1)(n-2)\dotsm 1,$$ we have $$5! = 120.$$ Hence the number of words beginning with $$E$$ or $$H$$ is $$2 \times 5! = 2 \times 120 = 240.$$ So far, $$\text{count} = 240.$$

Second position: We now fix the first letter as $$M$$ and look at the second letter, which in ‘MOTHER’ is $$O.$$ The remaining unused letters are $$E, H, O, R, T.$$ Among these, the letters alphabetically smaller than $$O$$ (while $$M$$ is already used) are $$E$$ and $$H,$$ i.e. two letters. With any one of these two as the second letter, the last four positions can be arranged in $$4! = 24$$ ways. Hence the additional words are $$2 \times 4! = 2 \times 24 = 48.$$ Adding to the previous total, $$\text{count} = 240 + 48 = 288.$$

Third position: We now have the prefix $$MO.$$ The third letter in ‘MOTHER’ is $$T.$$ The remaining unused letters are $$E, H, R, T.$$ Alphabetically smaller letters than $$T$$ within these are $$E, H, R,$$ i.e. three letters. Once any one of these three is chosen for the third position, the last three positions can be filled in $$3! = 6$$ ways. Thus the extra words contributed here are $$3 \times 3! = 3 \times 6 = 18.$$ Adding, $$\text{count} = 288 + 18 = 306.$$

Fourth position: The current prefix is $$MOT.$$ The fourth letter of ‘MOTHER’ is $$H.$$ The unused letters now are $$E, H, R.$$ The letter alphabetically smaller than $$H$$ in this set is only $$E.$$ That is just one letter. After fixing this $$E$$ as the fourth letter, the remaining two positions can be arranged in $$2! = 2$$ ways. Therefore the contribution is $$1 \times 2! = 1 \times 2 = 2.$$ Update the total: $$\text{count} = 306 + 2 = 308.$$

Fifth position: Our prefix is $$MOTH.$$ The fifth letter of ‘MOTHER’ is $$E.$$ The two unused letters are $$E$$ and $$R,$$ but $$E$$ is the smallest among them. There is no letter smaller than $$E$$ left, so no new words are added at this step.

Sixth position: The final letter $$R$$ is forced, and again no additional words arise.

Thus, exactly $$308$$ words precede ‘MOTHER’ in dictionary order. Hence the position of ‘MOTHER’ itself is $$308 + 1 = 309.$$

So, the answer is $$309$$.

First, let us analyse the word EXAMINATION. It contains 11 letters in all, but several of them repeat. Listing each distinct letter with its frequency, we have

$$E_1,\; X_1,\; A_2,\; M_1,\; I_2,\; N_2,\; T_1,\; O_1$$

Thus the eight different letters and their available copies are

$$E(1),\; X(1),\; A(2),\; M(1),\; I(2),\; N(2),\; T(1),\; O(1).$$

We wish to form all possible 4-letter “words” (ordered arrangements) without exceeding these individual limits. Order matters, so the task is to count all 4-permutations drawn from this multiset.

Because no letter occurs more than twice in the original word, the possible repetition patterns inside any 4-letter word are only

$$\;(1,1,1,1),\quad (2,1,1),\quad (2,2).$$

Patterns like $$(3,1)$$ or $$(4)$$ are impossible, since no letter appears three or four times in the source.

We now treat each admissible pattern separately and add the results.

Case I : all four letters different $$(1,1,1,1)$$

We first choose 4 distinct letters out of the 8 available. Using the combination formula $$\binom{n}{r}= \dfrac{n!}{r!\,(n-r)!},$$ we get

$$\binom{8}{4}=70.$$ Once the 4 distinct letters are chosen, they can be arranged in $$4! = 24$$ different ways. Hence

$$\text{words in this case}=70 \times 24 = 1680.$$

Case II : one letter repeated twice and two other different letters $$(2,1,1)$$

(a) Choosing the letter that appears twice: only the letters A, I and N possess two copies, so we have $$3\ \text{choices}.$$

(b) After fixing that repeated letter, 7 other distinct letters remain (because 1 of the 8 has already been used). We must pick any 2 of these, so

$$\binom{7}{2}=21.$$ (c) For a multiset containing two identical and two different letters, the permutation count is obtained by the division rule: $$\frac{4!}{2!}=12.$$

Therefore

$$\text{words in this case}=3 \times 21 \times 12 = 756.$$

Case III : two letters, each appearing twice $$(2,2)$$

(a) Both repeated letters must again come from the set {A, I, N}. Selecting any 2 of them gives

$$\binom{3}{2}=3.$$ (b) A multiset of the form $$a,a,b,b$$ has $$\frac{4!}{2!\,2!}=6$$ distinct permutations.

Hence

$$\text{words in this case}=3 \times 6 = 18.$$

Adding all cases together

$$\begin{aligned} \text{Total words} &= 1680 + 756 + 18 \\ &= 2454. \end{aligned}$$

So, the answer is $$2454$$.

First, we list the letters in the word ‘SYLLABUS’. We have $$S,\,Y,\,L,\,L,\,A,\,B,\,U,\,S$$. So, the multiset of letters contains $$S$$ twice, $$L$$ twice, and $$Y,\,A,\,B,\,U$$ each once.

We want to form 4-letter words in which exactly two letters are alike (a pair) and the remaining two letters are different from each other and from the repeated letter. Symbolically we need a pattern of the type $$a\,a\,b\,c$$ with $$b\neq c\neq a$$.

Step 1: choose the letter that appears twice. Only letters that occur at least twice in ‘SYLLABUS’ can serve this role. Those are $$S$$ and $$L$$. Hence, the repeated letter can be chosen in $$2$$ ways.

Step 2: choose the two distinct letters $$b$$ and $$c$$ from the remaining letters, making sure they are different from each other and from the repeated letter.

If the repeated letter is $$S$$, the available distinct types are $$\{Y,\,L,\,A,\,B,\,U\}$$, i.e. 5 different letters. The number of ways to pick any two of them is the combination $$\binom{5}{2}=10.$$

If the repeated letter is $$L$$, the remaining distinct types are $$\{S,\,Y,\,A,\,B,\,U\}$$, again 5 letters, giving $$\binom{5}{2}=10$$ ways.

So, the total number of choices for the 4-letter multiset $$\{a,a,b,c\}$$ is $$2\times 10 = 20.$$

Step 3: arrange the chosen letters. For any specific selection $$a,a,b,c$$, we must count all permutations of these four symbols where the two $$a$$’s are indistinguishable but $$b$$ and $$c$$ are distinguishable. The permutation formula for a multiset says $$\text{Number of permutations}=\dfrac{4!}{2!}=12,$$ because we divide by $$2!$$ to account for the identical pair of $$a$$’s.

Step 4: multiply the number of selections by the number of arrangements: $$20 \times 12 = 240.$$

So, the answer is $$240$$.

The given word is ‘LETTER’.  We first note the individual letters and how many times each appears:

$$L : 1,\; E : 2,\; T : 2,\; R : 1.$$

We want to count all possible arrangements (words) of these six letters in which the two vowels (both $$E$$’s) are never next to each other.

We begin by finding the total number of arrangements of all six letters without any restriction. The general formula for permutations of $$n$$ objects where some objects repeat is stated first:

$$\text{If } n_1, n_2, \ldots, n_k$$ are the repetition counts, then $$\dfrac{n!}{n_1!\,n_2!\,\ldots\,n_k!}.$$

Here, out of the six letters, the letter $$E$$ repeats $$2$$ times and the letter $$T$$ repeats $$2$$ times. Therefore,

Total unrestricted arrangements $$= \dfrac{6!}{2!\,2!}.$$

Using the factorial definition $$n! = n \times (n-1) \times \cdots \times 1,$$ we have

$$6! = 720,\qquad 2! = 2.$$

Substituting, we get

$$\dfrac{6!}{2!\,2!}= \dfrac{720}{2 \times 2}= \dfrac{720}{4}=180.$$

Now we must subtract those arrangements in which the two vowels come together. To enforce “togetherness,” we tie the two $$E$$’s into a single super-letter $$\bigl(EE\bigr).$$ Treating this super-letter as one entity, the objects we must arrange are

$$(EE),\,L,\,T,\,T,\,R.$$

That is a total of $$5$$ objects, with $$T$$ repeating twice. Using the same repetition formula, the number of arrangements with the two vowels side by side is

$$\dfrac{5!}{2!}.$$

Again, evaluate:

$$5! = 120,\qquad 2! = 2,$$

so

$$\dfrac{5!}{2!}= \dfrac{120}{2}=60.$$

Finally, to get the arrangements in which the vowels are never together, we subtract the “together” count from the total:

$$\text{Required number}= 180 - 60 = 120.$$

So, the answer is $$120$$.

We are looking for all three-digit natural numbers. Such a number can be written in decimal form as $$\overline{abc}$$ where $$a$$ is the hundreds digit, $$b$$ is the tens digit and $$c$$ is the units digit.

Because the number is a three-digit number, the leading digit $$a$$ cannot be zero, so we must have $$1 \le a \le 9$$. The other two digits can be anything from $$0$$ to $$9$$, so $$0 \le b \le 9$$ and $$0 \le c \le 9$$.

We are told that the sum of the three digits equals $$10$$. Hence we need every integer solution of the linear equation

$$a + b + c = 10,$$

subject to

$$1 \le a \le 9,\quad 0 \le b \le 9,\quad 0 \le c \le 9.$$

To remove the lower bound on $$a$$, we make the substitution $$a' = a - 1$$. Because $$a$$ ranges from $$1$$ to $$9$$, the new variable $$a'$$ ranges from $$0$$ to $$8$$. In terms of $$a'$$, the equation becomes

$$\bigl(a' + 1\bigr) + b + c = 10.$$

Simplifying, we obtain

$$a' + b + c = 9,$$

with the simpler bounds

$$0 \le a' \le 8,\quad 0 \le b \le 9,\quad 0 \le c \le 9.$$

Now we must count the number of non-negative integer solutions of $$a' + b + c = 9$$ that also obey the upper limits $$a' \le 8,\; b \le 9,\; c \le 9$$. Observe that the total we seek is small enough that the natural upper limits $$b \le 9$$ and $$c \le 9$$ are automatically respected once the sum is only $$9$$, so the only possible overflow is $$a' = 9$$. Therefore we first count all non-negative solutions and then subtract the ones in which $$a' = 9$$ (because in that case $$a'$$ would exceed $$8$$).

The standard stars-and-bars result says: “The number of non-negative integer solutions of $$x_1 + x_2 + x_3 = n$$ is $$\binom{n+2}{2}$$.” We state it explicitly here before using it.

Applying the formula with $$n = 9$$ gives

$$\text{Total unrestricted solutions} \;=\; \binom{9+2}{2} = \binom{11}{2} = \frac{11 \times 10}{2} = 55.$$

Now we exclude the disallowed case $$a' = 9$$. Fixing $$a' = 9$$ forces $$b + c = 0,$$ so we must have $$b = 0$$ and $$c = 0$$. That is exactly one solution. No other violation is possible because if $$a'$$ were $$10$$ or more, the sum would exceed $$9$$.

Therefore the number of admissible solutions is

$$55 - 1 = 54.$$

Each admissible triple $$(a', b, c)$$ corresponds uniquely to $$a = a' + 1$$, so we have counted every valid three-digit number whose digits add to $$10$$.

So, the answer is $$54$$.

The urn has $$5$$ red, $$4$$ black and $$3$$ white marbles, so the total number of marbles is $$5+4+3=12$$.

We must select $$4$$ marbles. When every marble is distinguishable, the basic counting rule for choosing $$r$$ objects from $$n$$ distinct objects is the combination formula

$$^nC_r=\dfrac{n!}{r!(n-r)!}.$$

Applying this formula to the complete set of $$12$$ marbles, the total number of possible samples of $$4$$ marbles is

$$^{12}C_4=\dfrac{12!}{4!\,8!}=495.$$

However, the condition “at the most three of them are red” forbids exactly one case: all four chosen marbles being red. We therefore compute how many selections consist of four red marbles and then subtract those unwanted selections from the total.

Among the $$5$$ red marbles, the number of ways to choose $$4$$ red marbles is

$$^{5}C_4=\dfrac{5!}{4!\,1!}=5.$$

No other selection violates the condition, because any mixture containing $$0,1,2$$ or $$3$$ red marbles is allowed. Thus the required number of favourable selections is obtained by excluding the $$5$$ completely red selections from the total $$495$$ selections:

$$\text{Favourable ways}=^{12}C_4-^{5}C_4=495-5=490.$$

So, the answer is $$490$$.

We have a total of 8 males and 5 females, so altogether $$8+5=13$$ people. A committee must have 11 members.

Remember the basic combination formula: to choose $$r$$ objects from $$n$$ distinct objects, the number of ways is given by $$^nC_r=\dfrac{n!}{r!\,(n-r)!}$$.

First we find $$m$$, the number of committees that contain at least 6 males. “At least 6” means 6 or 7 or 8 males can be in the committee. Because the committee has 11 members in total, once we decide the number of males, the number of females is automatically fixed.

For 6 males we need $$11-6=5$$ females. The number of ways is $$ ^8C_6 \times ^5C_5 =28\times1 =28. $$

For 7 males we need $$11-7=4$$ females. The number of ways is $$ ^8C_7 \times ^5C_4 =8\times5 =40. $$

For 8 males we need $$11-8=3$$ females. The number of ways is $$ ^8C_8 \times ^5C_3 =1\times10 =10. $$

Adding all these disjoint possibilities, $$ m = 28 + 40 + 10 = 78. $$

Now we find $$n$$, the number of committees that contain at least 3 females. “At least 3” means 3, 4, or 5 females can be in the committee. Once again the total remains 11.

For 3 females we need $$11-3=8$$ males. The number of ways is $$ ^5C_3 \times ^8C_8 =10\times1 =10. $$

For 4 females we need $$11-4=7$$ males. The number of ways is $$ ^5C_4 \times ^8C_7 =5\times8 =40. $$

For 5 females we need $$11-5=6$$ males. The number of ways is $$ ^5C_5 \times ^8C_6 =1\times28 =28. $$

Adding these possibilities, $$ n = 10 + 40 + 28 = 78. $$

Thus we have $$m = n = 78$$.

Hence, the correct answer is Option C.

We have a total of 5 boys and $$n$$ girls in the group, so the total number of students is $$5+n$$. We need to form teams of exactly 3 students, with the extra condition that each chosen team must contain at least one boy and at least one girl. This requirement can be met in only two mutually exclusive ways:

• the team has 1 boy and 2 girls, or
• the team has 2 boys and 1 girl.

We now count the number of selections for each of these two cases using the combination formula. First, we recall the formula for combinations:

$$^rC_s \;=\; \binom{r}{s} \;=\; \dfrac{r!}{s!\,(r-s)!}\;,$$

which gives the number of ways to choose $$s$$ objects out of $$r$$ distinct objects when order does not matter.

Case 1 (1 boy, 2 girls):
We select 1 boy out of 5 and 2 girls out of $$n$$. The count is therefore

$$\binom{5}{1}\,\binom{n}{2}.$$

Case 2 (2 boys, 1 girl):
We select 2 boys out of 5 and 1 girl out of $$n$$. The count is

$$\binom{5}{2}\,\binom{n}{1}.$$

Since the two cases are disjoint and together exhaust all possibilities that meet the given condition, the total number of valid teams is the sum of the two counts. According to the statement of the problem, this total equals 1750, so we write

$$\binom{5}{1}\binom{n}{2} \;+\; \binom{5}{2}\binom{n}{1} \;=\; 1750.$$

We now substitute the numerical values of the combinations involving only the boys:

First, $$\binom{5}{1}=5,$$ and $$\binom{5}{2}=\dfrac{5\cdot4}{2\cdot1}=10.$$

Hence the equation becomes

$$5\binom{n}{2}+10\binom{n}{1}=1750.$$

Next, we express the remaining combinations in algebraic form. For any positive integer $$n$$, we have

$$\binom{n}{2}=\dfrac{n(n-1)}{2},\qquad \binom{n}{1}=n.$$

Substituting these expressions into our equation, we obtain

$$5\left(\dfrac{n(n-1)}{2}\right)+10(n)=1750.$$

We now perform each algebraic step carefully. First multiply the fraction inside the parentheses by 5:

$$\dfrac{5}{2}\bigl(n^2-n\bigr)+10n=1750.$$

To clear the denominator, multiply every term of the equation by 2:

$$5\bigl(n^2-n\bigr)+20n=3500.$$

Next, distribute the 5 inside the parentheses:

$$5n^2-5n+20n=3500.$$

Combine the like terms $$-5n+20n$$ to get $$15n$$:

$$5n^2+15n=3500.$$

Divide the entire equation by 5 to simplify:

$$n^2+3n=700.$$

Now bring 700 to the left-hand side to form a quadratic equation equal to zero:

$$n^2+3n-700=0.$$

We solve this quadratic using the quadratic formula. For an equation of the form $$ax^2+bx+c=0$$, the solutions are

$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

Here, $$a=1,\; b=3,\; c=-700$$. Substituting these values, we get

$$n=\dfrac{-3\pm\sqrt{3^2-4\cdot1\cdot(-700)}}{2\cdot1}.$$

Simplify inside the square root:

$$3^2=9,\quad -4\cdot1\cdot(-700)=+2800,$$

so

$$n=\dfrac{-3\pm\sqrt{9+2800}}{2}=\dfrac{-3\pm\sqrt{2809}}{2}.$$

Compute the square root: $$\sqrt{2809}=53$$ because $$53^2=2809.$$ Thus

$$n=\dfrac{-3\pm53}{2}.$$

This gives two possible values:

1. $$n=\dfrac{-3+53}{2}=\dfrac{50}{2}=25,$$
2. $$n=\dfrac{-3-53}{2}=\dfrac{-56}{2}=-28.$$

The number of girls cannot be negative, so we discard $$n=-28$$. Therefore, the only acceptable value is

$$n=25.$$

The option list shows that 25 corresponds to Option C.

Hence, the correct answer is Option C.

We have the multiset of digits $$\{1,1,2,2,2,2,3,4,4\}$$ which contains a total of 9 digits.

The odd digits are $$1,1,3$$ — that is, two 1’s and one 3, giving exactly 3 odd digits. The even digits are $$2,2,2,2,4,4$$ — four 2’s and two 4’s, giving 6 even digits.

In any 9-digit number, the positions are numbered $$1,2,3,4,5,6,7,8,9$$. Even positions are $$2,4,6,8$$ (four places) and odd positions are $$1,3,5,7,9$$ (five places).

The requirement is: all odd digits must be placed only in even positions. Because there are only 3 odd digits but 4 even places, we must choose exactly 3 of those 4 even places for the odd digits, leaving the remaining one even place plus all five odd places for even digits.

First, we choose the positions for the odd digits. The number of ways to choose 3 places out of 4 is given by the combination formula $$\binom{n}{r} = \frac{n!}{r!\,(n-r)!}.$$ Substituting $$n=4,\; r=3$$ we get $$\binom{4}{3} = \frac{4!}{3!\,1!} = 4.$$

Next, we arrange the odd digits $$1,1,3$$ in the 3 chosen places. The number of distinct permutations of 3 objects in which two are identical is $$\frac{3!}{2!} = 3$$ because we divide by $$2!$$ to account for the repetition of the two 1’s.

Hence, the total number of ways to place the odd digits is $$4 \times 3 = 12.$$

Now we place the 6 even digits $$2,2,2,2,4,4$$ in the remaining 6 positions (five odd positions and the one unused even position). The number of distinct permutations of these 6 digits, with four 2’s and two 4’s, is $$\frac{6!}{4!\,2!} = \frac{720}{24 \times 2} = 15.$$

Finally, by the multiplication principle, the overall number of admissible 9-digit numbers is $$12 \times 15 = 180.$$

Hence, the correct answer is Option C.

We have a total of 5 girls and 7 boys in the class. Among the boys, two particular boys are named A and B, and they do not want to be in the same team. We want teams that always contain exactly 2 girls and 3 boys.

We first count the number of teams without any restriction and then subtract the “bad” teams that contain both A and B.

The combination formula is stated as: $$ ^nC_r \;=\; \frac{n!}{r!\,(n-r)!} $$ which gives the number of ways to choose $$r$$ objects out of $$n$$ objects.

Unrestricted count: For girls, we need 2 out of 5. $$ \text{Ways for girls} \;=\; ^5C_2 $$ Now, by direct substitution, $$ ^5C_2 \;=\; \frac{5!}{2!\,3!} \;=\; \frac{120}{2 \times 6} \;=\; 10. $$

For boys, we need 3 out of 7. $$ \text{Ways for boys} \;=\; ^7C_3 $$ Again using the formula, $$ ^7C_3 \;=\; \frac{7!}{3!\,4!} \;=\; \frac{5040}{6 \times 24} \;=\; 35. $$

The Product Rule says the total number of independent choices equals the product of the individual numbers of ways. So, the total number of unrestricted teams is $$ 10 \times 35 \;=\; 350. $$

Now we remove the “bad” teams in which both special boys A and B appear together.

Fixing A and B in the team leaves room for only one more boy. The remaining boys are 7 − 2 = 5 in number.

The number of ways to choose that one additional boy is $$ ^5C_1 = 5. $$

The girls are still chosen in the unrestricted way, namely $$ ^5C_2 = 10. $$

So, the number of “bad” teams containing both A and B is $$ 10 \times 5 \;=\; 50. $$

Finally, by the Subtraction Principle, the required number of valid teams is $$ 350 \;-\; 50 \;=\; 300. $$

Hence, the correct answer is Option A.

First, recall the basic counting fact: if a finite set has $$n$$ elements, the total number of all its subsets (including the empty set) is $$2^{\,n}$$, because each element may be either “chosen” or “not chosen”.

Here the set is $$S=\{1,2,3,\ldots ,100\}$$, so $$|S| = 100$$. Hence, by the above fact, the total number of subsets of $$S$$ is $$2^{100}$$. Among these, exactly one subset is empty, so the number of non-empty subsets is

$$2^{100}-1.$$

We are asked to count those non-empty subsets $$A$$ whose product of elements is even. A product of integers is even as soon as at least one factor is even. Therefore a subset has an even product precisely when it contains at least one even number. The complementary situation is that the subset contains no even numbers at all, i.e. it is formed entirely of odd numbers; such a subset will clearly have an odd product.

So, instead of counting the “even-product” subsets directly, we count all non-empty subsets and subtract the non-empty subsets that consist only of odd numbers. This is the classic “complement counting” technique.

Among the first 100 positive integers, exactly half are odd and half are even, because the list alternates parity. Concretely, the odd numbers are

$$1,3,5,\ldots ,99,$$

and there are $$50$$ of them. Let us call this set of odd numbers $$O$$, so $$|O| = 50$$.

Applying again the subset-counting fact, the total number of subsets of $$O$$ (including the empty subset) is $$2^{50}$$. Excluding the empty subset, the number of non-empty subsets of $$O$$ is therefore

$$2^{50}-1.$$

Every such subset uses only odd elements, hence its product is odd. Consequently, the count $$2^{50}-1$$ is exactly the number of non-empty subsets of $$S$$ having an odd product.

Now we subtract these from the total number of non-empty subsets to obtain the desired count of even-product subsets:

$$ \begin{aligned} \text{Number of even-product subsets} &= (2^{100}-1) \;-\; (2^{50}-1)\\[4pt] &= 2^{100}-1-2^{50}+1\\[4pt] &= 2^{100}-2^{50}.\\ \end{aligned} $$

This difference can be factorised by taking $$2^{50}$$ common:

$$ 2^{100}-2^{50} = 2^{50}\left(2^{100-50}-1\right) = 2^{50}\left(2^{50}-1\right). $$

Thus the number of non-empty subsets of $$S$$ whose product is even is $$2^{50}\bigl(2^{50}-1\bigr)$$.

Looking at the given options, this expression matches Option C.

Hence, the correct answer is Option C.

We have a total of $$n = 20$$ identical pillars placed at equal intervals on the circumference of a circular stadium.

Any pair of pillars can potentially be connected by a beam. The number of different pairs that can be chosen from $$n$$ objects is given by the combination formula

$$^nC_2 \;=\; \frac{n(n-1)}{2}.$$

Substituting $$n = 20$$, we get

$$^{20}C_2 \;=\; \frac{20 \times 19}{2} \;=\; 190.$$

These 190 pairs represent all possible connections, including the connections between adjacent pillars (the sides of the 20-gon) and the connections between non-adjacent pillars (the chords of the 20-gon).

However, the problem states that beams are laid only between non-adjacent pillars. We must therefore exclude those pairs which are adjacent.

In a 20-sided polygon, each side joins exactly two adjacent pillars, and there are as many sides as pillars. Hence the number of adjacent pairs equals

$$20.$$

So the number of beams required is obtained by subtracting these 20 adjacent connections from the total 190 possible connections:

$$\text{Required beams} \;=\; 190 \;-\; 20 \;=\; 170.$$

Therefore, the total number of beams that can be fixed between the tops of all non-adjacent pillars is $$170$$. Hence, the correct answer is Option A.

We have to form six-digit numbers with the six distinct digits 0, 1, 2, 5, 7 and 9. Because every digit is used exactly once, each admissible arrangement is a permutation of this complete set, but only those permutations that satisfy the divisibility rule for 11 will be counted.

First, recall the test for divisibility by 11: a number is divisible by 11 if and only if the alternating sum of its digits is a multiple of 11. Writing a six-digit number as $$d_1d_2d_3d_4d_5d_6$$ (with $$d_1$$ the left-most digit), the rule states

$$\bigl(d_1+d_3+d_5\bigr)\;-\;\bigl(d_2+d_4+d_6\bigr)\equiv 0\pmod{11}.$$

Let us denote

$$S_{\text{odd}} \;=\; d_1+d_3+d_5,$$ $$S_{\text{even}} = d_2+d_4+d_6.$$

Since the six given digits are all used, their total sum is fixed:

$$0+1+2+5+7+9 \;=\; 24.$$

Thus

$$S_{\text{odd}}+S_{\text{even}} = 24.$$

To satisfy the 11-test we need

$$S_{\text{odd}}-S_{\text{even}}\equiv 0\pmod{11}.$$

The difference between two sums of three digits cannot exceed 18 in magnitude, so the only multiple of 11 lying in the interval $$[-18,18]$$ is 0 itself. Hence we must have

$$S_{\text{odd}}-S_{\text{even}} = 0,$$

which gives

$$S_{\text{odd}} = S_{\text{even}} = 12.$$

We now seek all unordered triples of the set {0,1,2,5,7,9} that add up to 12. Listing every possibility:

$$0+5+7 = 12,\qquad 1+2+9 = 12.$$

No other combination of three distinct digits from the set sums to 12, so the only way to split the six digits into two equal-sum groups is

$$\{0,5,7\}\quad\text{and}\quad\{1,2,9\}.$$

Either of these two triples can occupy the odd positions while the other occupies the even positions. Therefore there are exactly

$$2\;$$ ways to assign the two triples to the two position sets.

Once an assignment is fixed, the three chosen digits can be permuted among their three designated positions. Each set therefore contributes

$$3!\times 3! = 6\times 6 = 36$$

arrangements. Multiplying by the two possible assignments gives an initial count

$$2\times 36 = 72.$$

However, some of these 72 arrangements do not represent valid six-digit numbers because a number is not allowed to start with 0. Let us remove the invalid cases.

• If the triple $$\{0,5,7\}$$ occupies the odd positions, the left-most position $$d_1$$ can receive the digit 0. Fixing $$d_1=0,$$ the remaining two odd positions $$d_3,d_5$$ can be filled with the digits 5 and 7 in $$2! = 2$$ ways. The even positions admit $$3! = 6$$ permutations of the digits 1, 2 and 9. Consequently the number of inadmissible arrangements with a leading zero equals

$$2\times 6 = 12.$$

• If the triple $$\{1,2,9\}$$ occupies the odd positions, then $$d_1$$ is one of 1, 2 or 9, never 0, so no additional arrangements are excluded in this case.

Hence the total count of acceptable six-digit numbers is

$$72-12 = 60.$$

Therefore we can form 60 different six-digit numbers using the digits 0, 1, 2, 5, 7 and 9, each used once, that are divisible by 11.

Hence, the correct answer is Option B.

We have a total of 31 objects. 10 of these are identical (say identical balls) and the remaining 21 objects are all distinct. We wish to select exactly 10 objects altogether.

Let us suppose that out of the 10 identical objects we actually pick $$k$$ of them. Because they are identical, picking $$k$$ of them can be done in only one way; there is no further choice involved once the value of $$k$$ is fixed.

Having decided to take $$k$$ identical objects, we must still choose the remaining $$10-k$$ objects from the 21 distinct ones. The number of ways to choose $$10-k$$ objects out of 21 distinct objects is, by the definition of combinations,

$$\binom{21}{\,10-k\,} = \frac{21!}{(10-k)!\,(21-(10-k))!}.$$

Here $$k$$ can be any integer from 0 up to 10 (we cannot take more than 10 identical objects because we only need 10 in total). Therefore the total number of possible selections is obtained by summing the above expression over all admissible $$k$$:

$$\text{Total ways} \;=\; \sum_{k=0}^{10} \binom{21}{\,10-k\,}.$$

For convenience we perform a change of variable. Put $$r = 10-k$$. Then when $$k = 0$$ we have $$r = 10$$, and when $$k = 10$$ we have $$r = 0$$. Hence $$r$$ runs through all the integers from 0 to 10 inclusive, and the sum rewrites as

$$\text{Total ways} \;=\; \sum_{r=0}^{10} \binom{21}{\,r\,}.$$

Now we recall two standard binomial identities:

1. The Binomial-Theorem sum $$\displaystyle \sum_{r=0}^{n} \binom{n}{r} = 2^{n}.$$

2. The symmetry of binomial coefficients $$\displaystyle \binom{n}{r} = \binom{n}{\,n-r\,}.$$

For our case $$n = 21$$, an odd number. Listing all coefficients $$\binom{21}{0},\binom{21}{1},\dots,\binom{21}{21}$$, there are 22 terms in total. They come in pairs of equal value: $$\binom{21}{0} = \binom{21}{21},\,\binom{21}{1} = \binom{21}{20},$$ and so on, up to $$\binom{21}{10} = \binom{21}{11}.$$ Because 21 is odd, there is no single unpaired “middle” coefficient. Consequently, half of the full binomial sum is obtained by adding the first 11 terms (those with $$r=0$$ to $$r=10$$), and the other half is obtained by the remaining 11 terms (those with $$r=11$$ to $$r=21$$).

Hence we have

$$\sum_{r=0}^{10} \binom{21}{\,r\,} \;=\; \frac{1}{2}\,\sum_{r=0}^{21} \binom{21}{\,r\,} \;=\; \frac{1}{2}\,\bigl(2^{21}\bigr) \;=\; 2^{20}.$$

Therefore the number of ways to choose the required 10 objects is $$2^{20}.$$

Hence, the correct answer is Option A.

Let us denote the number of men by the symbol $$m$$. Besides these men there are exactly two women in the tournament, so the total number of participants is $$m+2$$.

By the condition of the tournament, every distinct pair of participants plays two games with each other. We must compare two separate counts:

1. The total number of games played among the men themselves.

2. The total number of games played between the men and the women.

To find the first count, we start with the well-known combination formula. The number of unordered pairs that can be chosen from $$m$$ men is

$$\binom{m}{2}=\frac{m(m-1)}{2}.$$

For each such pair, exactly two games are played. Therefore, the number of games only among the men is

$$\text{Games among men}=2 \times \binom{m}{2}=2 \times \frac{m(m-1)}{2}=m(m-1).$$

Next, we consider the games between men and women. There are $$m$$ ways to pick a man and $$2$$ ways to pick a woman, giving $$m \times 2 = 2m$$ distinct man-woman pairs. Each such pair also plays two games, so

$$\text{Games between men and women}=2 \times (2m)=4m.$$

The problem states that the number of games among the men exceeds the number of games between men and women by $$84$$. Translating this directly into an equation, we have

$$m(m-1)-4m=84.$$

Now we simplify step by step:

First expand the left-hand side:

$$m(m-1)=m^2-m,$$

so

$$m^2-m-4m=84.$$

Combine the like terms $$-m-4m$$ to get $$-5m$$:

$$m^2-5m=84.$$

Bring all terms to one side to obtain a standard quadratic equation:

$$m^2-5m-84=0.$$

We now solve this quadratic by the quadratic formula. For an equation of the form $$ax^2+bx+c=0$$ the solutions are

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

Here we have $$a=1,\; b=-5,\; c=-84$$. Substituting these values, we get

$$m=\frac{-(-5)\pm\sqrt{(-5)^2-4\cdot1\cdot(-84)}}{2\cdot1}.$$

Simplify each part carefully:

$$-(-5)=5,$$

$$(-5)^2=25,$$

$$-4\cdot1\cdot(-84)=336,$$

so inside the square root we have

$$25+336=361.$$

Therefore,

$$m=\frac{5\pm\sqrt{361}}{2}.$$

Since $$\sqrt{361}=19$$, we obtain

$$m=\frac{5\pm19}{2}.$$

This gives two numerical possibilities:

$$m=\frac{5+19}{2}=24/2=12,$$

or

$$m=\frac{5-19}{2}=-14/2=-7.$$

The value $$m=-7$$ has no meaning in this context because a negative number of men is impossible. Hence the only permissible solution is

$$m=12.$$

Among the given options, $$12$$ corresponds to Option B.

Hence, the correct answer is Option B.

We have three separate boxes, and each box contains the ten balls whose labels are the integers $$1,2,3,\ldots ,10$$. From every box exactly one ball is drawn, so we obtain an ordered triple of labels $$(n_1,n_2,n_3)$$ where $$n_1$$ comes from the first box, $$n_2$$ from the second, and $$n_3$$ from the third.

The condition given in the question is $$n_1 \lt n_2 \lt n_3$$. Because of the two “<” symbols, the three numbers must be pairwise distinct; no two of them can be equal.

Since each box possesses every label from 1 to 10, once we decide which number is to be $$n_1$$, which is to be $$n_2$$, and which is to be $$n_3$$, there is exactly one way to realise that decision: we simply pick the ball carrying that label from the corresponding box. Thus the whole counting problem reduces to selecting the three distinct labels themselves.

So we only have to count the number of different subsets of three distinct numbers that can be taken from the set $$\{1,2,3,\ldots ,10\}$$. After choosing the three numbers, we automatically arrange them in increasing order to become $$n_1,n_2,n_3$$, so no further permutations are possible or needed.

To count such subsets we use the combination formula. The number of ways of choosing $$r$$ objects out of $$n$$ distinct objects is given by $$ \binom{n}{r}=\frac{n!}{r!\,(n-r)!}. $$ Here $$n=10$$ and $$r=3$$, hence $$ \binom{10}{3}= \frac{10!}{3!\,(10-3)!}= \frac{10\times9\times8\times7!}{3\times2\times1\times7!}= \frac{720}{6}=120. $$

Therefore there are $$120$$ favourable triples $$(n_1,n_2,n_3)$$ satisfying $$n_1 \lt n_2 \lt n_3$$.

Hence, the correct answer is Option C.