JEE Permutations & Combinations Questions

We need to form a group of 8 people (4 boys, 4 girls) such that 5 are from Group A (7B, 3G) and 3 are from Group B (6B, 5G).

First, we will distribute boys and girls from Group A.

Let Boys from A = x and Girls from A = y

Then, x + y = 5

Now, from Group B:

Boys = 4 - x

Girls = 4 - y

Now, the possible cases are:

Case 1: x = 4, y = 1

From A: $$^7C_4\times^3C_1=35\times3=105$$

From B: $$^6C_0\times^5C_3=1\times10=10$$

Total = 105*10 = 1050

Case 2: x = 3, y = 2

From A: $$^7C_3\times^3C_2=35\times3=105$$

From B: $$^6C_1\times^5C_2=6\times10=60$$

Total = 105*60 = 6300

Case 3: x = 2, y = 3

From A: $$^7C_2\times^3C_3=21\times1=21$$

From B: $$^6C_2\times^5C_1=15\times5=75$$

Total = 21*75 = 1575

Sum = 1050 + 6300 + 1575 = 8925

To find the highest power of a prime $$p$$ that divides $$m!$$, we use Legendre’s formula:
$$\nu_p(m!) = \left\lfloor\frac{m}{p}\right\rfloor + \left\lfloor\frac{m}{p^2}\right\rfloor + \left\lfloor\frac{m}{p^3}\right\rfloor + \cdots$$
where $$\nu_p(m!)$$ denotes the exponent of $$p$$ in the prime-factorisation of $$m!$$.

Here, $$p = 3$$ and $$m = 50$$. We compute each term until the quotient becomes zero:

First term: $$\left\lfloor\frac{50}{3}\right\rfloor = 16$$

Second term: $$\left\lfloor\frac{50}{9}\right\rfloor = 5$$

Third term: $$\left\lfloor\frac{50}{27}\right\rfloor = 1$$

Fourth term: $$\left\lfloor\frac{50}{81}\right\rfloor = 0$$ (and all further terms are also $$0$$)

Add the non-zero contributions:
$$\nu_3(50!) = 16 + 5 + 1 = 22$$

Therefore the greatest integer $$n$$ such that $$3^n$$ divides $$50!$$ is $$n = 22$$.

Hence, the correct option is Option B.

We have to select 5 alphabets and arrange them in alphabetical order such that the middle term is M. 

M is the 13th term in the English alphabets. 

It means that we have to select 2 alphabets out of 12, and place it on the left hand side of M and 2 alphabets out of the remaining 13, and place it on the right hand side of M. 

For the left hand side of M: 

Number of ways in which we select 2 alphabets out of 12 = $$12_{C_2}$$

Number of ways in which we arrange the selected alphabets in alphabetical order = 1 (only one case will be possible) 

Total number of ways in which we select 2 alphabets out of 12 and arrange it in alphabetical order = $$12_{C_2}\times1=12_{C_2}=66$$

For the right hand side of M: 

Number of ways in which we select 2 alphabets out of 13 = $$13_{C_2}$$

Number of ways in which we arrange the selected alphabets in alphabetical order = 1 (only one case will be possible)

Total number of ways in which we select 2 alphabets out of 13 and arrange it in alphabetical order = $$13_{C_2}\times1=13_{C_2}=78$$

Number of ways in which we can select M and place it in the middle of 5 alphabets = 1 

Total number of ways in which we select 5 alphabets and arrange it in alphabetical order such that M is the middle term = $$66\times78\times1=5148$$

Hence, we can select 5 alphabets out of 26 and arrange it in alphabetical order in such a way that the middle term is M in 5148 ways. 

$$\therefore\ $$ The required answer is A.

Let a seven-digit number $$x$$ contain $$a_0$$ zeros, $$a_1$$ ones and $$a_2$$ twos, so that

$$a_0 + a_1 + a_2 = 7$$ and $$a_0,\,a_1,\,a_2 \ge 0$$.

Because the first (left-most) digit cannot be $$0$$, we count the admissible arrangements for each triple $$(a_0,a_1,a_2)$$ as follows.

Total arrangements with these counts = $$\dfrac{7!}{a_0!\,a_1!\,a_2!}$$.
Arrangements whose first digit is $$0$$: fix a zero at the first place and permute the remaining $$6$$ digits, giving $$\dfrac{6!}{(a_0-1)!\,a_1!\,a_2!}$$ (this term appears only when $$a_0\ge 1$$).

Hence the required number for the triple $$(a_0,a_1,a_2)$$ is

$$N(a_0,a_1,a_2)= \begin{cases} \dfrac{7!}{a_0!\,a_1!\,a_2!}-\dfrac{6!}{(a_0-1)!\,a_1!\,a_2!}, & a_0\ge 1\\[6pt] \dfrac{7!}{a_0!\,a_1!\,a_2!}, & a_0=0 \end{cases}$$

For $$a_0\ge 1$$, simplifying gives

$$N(a_0,a_1,a_2)=\dfrac{6!}{a_0!\,a_1!\,a_2!}\,(7-a_0)\qquad -(1)$$

The problem asks for the count of all numbers in which at least one of the digits $$0$$ or $$1$$ appears exactly twice. We split this into three disjoint cases.

Let the possible values of $$a_1$$ be $$0,1,3,4,5$$ (they must stay within $$0\le a_1\le 5$$). For each value we compute $$a_2=7-2-a_1$$ and use $$(1)$$.

$$\begin{aligned} a_1=0,&\;a_2=5:&\;N=\dfrac{6!\,(7-2)}{2!\,0!\,5!}=15\\ a_1=1,&\;a_2=4:&\;N=\dfrac{6!\,(7-2)}{2!\,1!\,4!}=75\\ a_1=3,&\;a_2=2:&\;N=\dfrac{6!\,(7-2)}{2!\,3!\,2!}=150\\ a_1=4,&\;a_2=1:&\;N=\dfrac{6!\,(7-2)}{2!\,4!\,1!}=75\\ a_1=5,&\;a_2=0:&\;N=\dfrac{6!\,(7-2)}{2!\,5!\,0!}=15 \end{aligned}$$

Total for Case 1 = $$15+75+150+75+15 = 330$$.

Now $$a_0$$ can be $$0,1,3,4,5$$. Compute $$a_2=7-a_0-2$$ and apply the appropriate formula.

$$\begin{aligned} a_0=0,&\;a_2=5:&\;N=\dfrac{7!}{0!\,2!\,5!}=21\\ a_0=1,&\;a_2=4:&\;N=\dfrac{6!\,(7-1)}{1!\,2!\,4!}=90\\ a_0=3,&\;a_2=2:&\;N=\dfrac{6!\,(7-3)}{3!\,2!\,2!}=120\\ a_0=4,&\;a_2=1:&\;N=\dfrac{6!\,(7-4)}{4!\,2!\,1!}=45\\ a_0=5,&\;a_2=0:&\;N=\dfrac{6!\,(7-5)}{5!\,2!\,0!}=6 \end{aligned}$$

Total for Case 2 = $$21+90+120+45+6 = 282$$.

Both digits $$0$$ and $$1$$ appear exactly twice.

$$N=\dfrac{6!\,(7-2)}{2!\,2!\,3!}=150$$

Summing all three cases:

$$330 + 282 + 150 = 762$$

Therefore, the required number of seven-digit numbers is
762.

We need to find how many words can be formed using all letters of "DAUGHTER" such that all vowels never come together.

DAUGHTER has 8 letters: D, A, U, G, H, T, E, R

Vowels: A, U, E (3 vowels)

Consonants: D, G, H, T, R (5 consonants)

All letters are distinct.

Total = $$8! = 40320$$

Treat the 3 vowels as a single unit. We then have 6 units (5 consonants + 1 vowel group) to arrange:

Arrangements = $$6! \times 3! = 720 \times 6 = 4320$$

(The $$3!$$ accounts for internal arrangements of the vowels within the group.)

Required count = Total - Vowels together = $$40320 - 4320 = 36000$$

The correct answer is Option 1: 36000.

Any triangle is completely determined by choosing its three vertices.
Therefore, start by counting all possible selections of three points from the 12 points.

Total unordered triples of points $$= {}^{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$$ $$-(1)$$

A triangle cannot be formed if all three chosen points lie on the same straight line. The only collinear points given are the 5 points on one common line; all other triples are non-collinear.

Number of degenerate (collinear) triples $$= {}^{5}C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$$ $$-(2)$$

Subtract these degenerate cases from the total in $$(1)$$ to obtain the required number of triangles:

Required triangles $$= 220 - 10 = 210$$

Hence, the total number of distinct triangles that can be formed is $$210$$. Option D is correct.

The number of triangles that can be formed from the vertices of a regular $$n$$-gon is the number of ways of choosing any $$3$$ vertices:
$$p = {}^{n}C_{3} = \frac{n(n-1)(n-2)}{6}$$

The number of quadrilaterals is the number of ways of choosing any $$4$$ vertices:
$$q = {}^{n}C_{4} = \frac{n(n-1)(n-2)(n-3)}{24}$$

We are told that
$$p + q = 126$$

Substituting the expressions for $$p$$ and $$q$$:
$$\frac{n(n-1)(n-2)}{6} + \frac{n(n-1)(n-2)(n-3)}{24} = 126$$

Multiply every term by $$24$$ to clear denominators:
$$4n(n-1)(n-2) + n(n-1)(n-2)(n-3) = 3024$$

Factor out $$n(n-1)(n-2)$$:
$$n(n-1)(n-2)\,\bigl[\,4 + (n-3)\bigr] = 3024$$

Simplify the bracket:
$$n(n-1)(n-2)(n+1) = 3024$$

Try successive integer values of $$n \ge 4$$ until the product equals $$3024$$.
Case 6: $$6\cdot5\cdot4\cdot7 = 840 \lt 3024$$
Case 7: $$7\cdot6\cdot5\cdot8 = 1680 \lt 3024$$
Case 8: $$8\cdot7\cdot6\cdot9 = 3024$$ — match found.

Hence $$n = 8$$.

The ellipse given is
$$\frac{x^2}{16} + \frac{y^2}{n} = 1 \quad\Longrightarrow\quad \frac{x^2}{16} + \frac{y^2}{8} = 1$$ because $$n = 8$$.

For an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a^2 \gt b^2$$, the eccentricity is
$$e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$$

Here $$a^2 = 16,\; b^2 = 8$$, so
$$e = \sqrt{1 - \frac{8}{16}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

Therefore the required eccentricity is $$\boxed{\dfrac{1}{\sqrt{2}}}$$, which corresponds to Option D.

To find the number of ways to place the letters A, B, C, D, and E into the 8 boxes such that no row remains empty, we first determine the number of ways to select 5 boxes from the grid under this constraint, and then arrange the 5 distinct letters within those selected boxes.

The grid consists of 8 boxes distributed across 3 horizontal rows with the configuration of 3 boxes in the top row, 2 boxes in the middle row, and 3 boxes in the bottom row:

$$\text{Row 1} = 3 \text{ boxes}, \quad \text{Row 2} = 2 \text{ boxes}, \quad \text{Row 3} = 3 \text{ boxes}$$

Let $$r_1, r_2,$$ and $$r_3$$ be the number of boxes selected from Row 1, Row 2, and Row 3 respectively. We need to choose a total of 5 boxes ($$r_1 + r_2 + r_3 = 5$$) such that no row is empty ($$r_1, r_2, r_3 \ge 1$$).

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Step 1: List the possible distributions $$(r_1, r_2, r_3)$$ of selecting the 5 boxes

Case 1: When 1 box is selected from the middle row ($$r_2 = 1$$)

The remaining 4 boxes must be chosen from the top and bottom rows ($$r_1 + r_3 = 4$$):

Distribution (3, 1, 1): $$\binom{3}{3} \times \binom{2}{1} \times \binom{3}{1} = 1 \times 2 \times 3 = 6 \text{ ways}$$

Distribution (2, 1, 2): $$\binom{3}{2} \times \binom{2}{1} \times \binom{3}{2} = 3 \times 2 \times 3 = 18 \text{ ways}$$

Distribution (1, 1, 3): $$\binom{3}{1} \times \binom{2}{1} \times \binom{3}{3} = 3 \times 2 \times 1 = 6 \text{ ways}$$

Case 2: When 2 boxes are selected from the middle row ($$r_2 = 2$$)

The remaining 3 boxes must be chosen from the top and bottom rows ($$r_1 + r_3 = 3$$):

Distribution (2, 2, 1): $$\binom{3}{2} \times \binom{2}{2} \times \binom{3}{1} = 3 \times 1 \times 3 = 9 \text{ ways}$$

Distribution (1, 2, 2): $$\binom{3}{1} \times \binom{2}{2} \times \binom{3}{2} = 3 \times 1 \times 3 = 9 \text{ ways}$$

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Step 2: Calculate the total number of ways to select the 5 boxes

Summing the ways from all valid distributions:

$$\text{Total box selections} = 6 + 18 + 6 + 9 + 9 = 48 \text{ ways}$$

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Step 3: Arrange the 5 distinct letters in the selected boxes

The 5 distinct letters (A, B, C, D, E) can be arranged in the 5 chosen boxes in $$5!$$ ways:

$$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \text{ ways}$$

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Step 4: Compute the final number of placements

$$\text{Total number of ways} = 48 \times 120 = 5760$$

Therefore, the total number of ways to place the letters is equal to 5760.

An equivalence relation on a set must satisfy three properties: reflexivity, symmetry, and transitivity. The number of equivalence relations on a set is equal to the number of partitions of that set, as each partition defines an equivalence class.

For the set $$S = \{1, 2, 3\}$$, we need to find all possible partitions. A partition divides the set into non-empty, disjoint subsets whose union is the entire set.

The possible partitions are:

1. Partition with one subset:
- The entire set: $$\{\{1, 2, 3\}\}$$
This corresponds to one equivalence relation where all elements are related to each other.

2. Partitions with two subsets:
- Subset 1: $$\{1\}$$, Subset 2: $$\{2, 3\}$$
- Subset 1: $$\{2\}$$, Subset 2: $$\{1, 3\}$$
- Subset 1: $$\{3\}$$, Subset 2: $$\{1, 2\}$$
Each of these partitions corresponds to an equivalence relation where elements in the same subset are equivalent. There are three such partitions.

3. Partition with three subsets:
- Each element in its own subset: $$\{\{1\}, \{2\}, \{3\}\}$$
This corresponds to the equivalence relation where each element is only related to itself.

Total number of partitions = 1 (one subset) + 3 (two subsets) + 1 (three subsets) = 5.

Each partition defines a unique equivalence relation. Since reflexivity requires that every element is related to itself (i.e., pairs like (1,1), (2,2), (3,3) must be present), no equivalence relation can be empty. Therefore, all 5 equivalence relations are non-empty.

The number of non-empty equivalence relations is 5.

The correct option is B. 5.

We need 5-digit numbers greater than 50000 using digits 0, 1, 2, 3, 4, 5, 6, 7 (repetition allowed), where the sum of the first and last digits is at most 8.

Since the number must be greater than 50000, the first digit $$a$$ must be 5, 6, or 7.

The middle three digits can each be any of the 8 digits (0 through 7), giving $$8^3 = 512$$ choices.

For the last digit $$e$$, we need $$a + e \leq 8$$ where $$e \in \{0, 1, 2, 3, 4, 5, 6, 7\}$$.

Case 1: $$a = 5$$: then $$e \leq 3$$, so $$e \in \{0, 1, 2, 3\}$$ — 4 choices.

Case 2: $$a = 6$$: then $$e \leq 2$$, so $$e \in \{0, 1, 2\}$$ — 3 choices.

Case 3: $$a = 7$$: then $$e \leq 1$$, so $$e \in \{0, 1\}$$ — 2 choices.

Total number of such 5-digit numbers = $$(4 + 3 + 2) \times 512 = 9 \times 512 = 4608$$

However, the question asks for numbers strictly greater than 50000. The number 50000 itself satisfies all conditions (first digit 5, last digit 0, sum = 5 ≤ 8), but 50000 is not greater than 50000. So we must exclude it.

Therefore, the required count = $$4608 - 1 = 4607$$

Let's assume all girls together as 1 block (G) and all boys together as another block (B).

So we have 2 blocks: G and B. They can be arranged in:  2! = 2 ways

Arrange the girls inside their block = 3! = 6 ways

Arrange boys inside their block = 4! = 24 ways

Now, we can subtract the cases where B1 and B2 are adjacent. 

If we assume B1 and B2 as one unit, then we have 3 units of boys. They can be arranged in 3! ways, and B1 and B2 can also arrange themselves in two ways. 

So, adjacent cases = 3! × 2 = 6 × 2 = 12

Valid arrangements: 24 - 12 = 12

Hence, the total number of ways: 2 × 6 × 12 = 144

S = set of first 10 primes. A = S ∪ P where P = all possible products of distinct elements of S. Find ordered pairs (x,y) where x∈S, y∈A, x|y.

A consists of S (the 10 primes) plus all products of 2 or more distinct primes from S. Total elements in A = $$2^{10} - 1 = 1023$$ (all non-empty subsets of S, each giving a unique product).

For each prime $$p_i \in S$$, count elements y ∈ A such that $$p_i | y$$.

y is a product of a non-empty subset of S. For $$p_i | y$$, the subset must contain $$p_i$$.

Number of such subsets = $$2^9 = 512$$ (choose any subset of the remaining 9 primes, including empty set, and include $$p_i$$).

Total ordered pairs

Each of the 10 primes in S contributes 512 pairs.

Total = $$10 \times 512 = 5120$$

The answer is 5120.

Let the digits be $$a, b, c$$ , where  $$a \ge 2$$, and $$b, c \ge 0$$, and all ≤ 9.

We want $$ a + b + c = 15 $$ with the number between 212 and 999.

Let, $$ a = 2$$

If $$a = 2$$ , then the number must be ≥ 212, so $$b \ge 1$$

And, $$ b + c = 13$$

Valid pairs are $$ (4,9),(5,8),(6,7),(7,6),(8,5),(9,4) $$

That’s 6 pairs, all satisfy $$ b \ge 1$$ So, 6 numbers are obtained in this case. 

Similarly, we fix for other values of a.

a = 3, b+c = 12

Possible choices: (3,9) to (9,3) → 7 numbers

a = 4 , b+c = 11

Possible choices: (2,9) to (9,2) → 8 numbers

a = 5 , b+c = 10

Possible choices: (1,9) to (9,1) → 9 numbers

a = 6 , b+c = 9

Possible choices: (0,9) to (9,0) → 10 numbers

a = 7 , b+c = 8

Possible choices: (0,8) to (8,0) → 9 numbers

a = 8 , b+c = 7

Possible choices: (0,7) to (7,0) → 8 numbers

a = 9 , b+c = 6

Possible choices: (0,6) to (6,0) → 7 numbers

So, the total possible choices of numbers: 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64

We form 6-letter words from letters of MATHS (5 distinct letters), with condition:

Any letter used must appear at least twice.

Possible repetition patterns for total 6 letters:

Case 1: (2+2+2)

Choose 3 letters from 5:
$$\binom{5}{3}=10$$

Arrange:
$$\frac{6!}{2!2!2!}=90$$

Total:
$$10\times90=900$$

Case 2: (4+2)

Choose letter repeated 4 times:
5

Choose letter repeated 2 times from remaining 4:
4

Arrange:
$$\frac{6!}{4!2!}=15$$

Total:
$$5\times4\times15=300$$

Case 3: (3+3)

Choose 2 letters from 5:
$$\binom{5}{2}=10$$

Arrange:
$$\frac{6!}{3!3!}=20$$

Total:
$$10\times20=200$$

Case 4: (2+2+1+1) ❌ not allowed

(two letters appear once)

Case 5: (6)

All same letter:

5

Total 900+300+200+5=1405

If all the boys sit together, then there will be 4 girls and 1 group of boys. 

Number of ways in which the boys can be arranged = 5!

Number of ways in which 4 girls and 1 group of boys can be arranged = 5!

Total number of ways in which we can arrange 4 girls and 5 boys such that all the boys sit together = $$5!\times5!=14400$$

Now, we have to arrange the boys and girls such that no two boys sit together. 

So, we will first fix the positions of girls in 1 way where girls are sitting at the alternate positions (not at the first or the last position) out of the total 9 positions.

Girls will sit at 2nd, 4th, 6th and 8th positions.  

Number of ways in which the girls are arranged = 4!

Now, the boys will sit in the remaining places.  

Number of ways in which the boys are arranged = 5! 

Total number of ways in which we can arrange the boys and girls such that no two boys sit together = $$4!\times5!=2880$$

Total number of ways in which we can arrange 5 boys and 4 girls such that either all the boys sit together or no boys sit together at all = $$14400+2880=17280$$

$$\therefore\ $$ The required answer is 17280.

Let $$S$$ be the set of all seven-digit natural numbers. A seven-digit number has its first (left-most) digit from $$1$$ to $$9$$ and each of the remaining six digits from $$0$$ to $$9$$, so

total elements in $$S = 9 \times 10^{6} = 90\,00\,00\,00 = 9{,}000{,}000$$

We must count those numbers in $$S$$ whose digit-sum is even. Denote this subset by $$E$$ and let $$O$$ be the subset with odd digit-sum.

Key idea (Parity-changing bijection): For every number in $$S$$ we will construct a unique partner in $$S$$ whose digit-sum parity is opposite.
Take a number with decimal representation $$a_1a_2a_3a_4a_5a_6a_7$$.

Define a mapping $$f:S \rightarrow S$$ by changing only the last digit:
• if $$a_7$$ is even, replace it by $$a_7+1$$ (this is odd);
• if $$a_7$$ is odd, replace it by $$a_7-1$$ (this is even).

This mapping leaves the first digit $$a_1$$ unchanged (so the image is still seven-digit) and clearly toggles the parity of the digit-sum. Moreover, the rule is reversible—applying it twice returns the original number—so $$f$$ is a bijection between $$E$$ and $$O$$.

Therefore $$|E| = |O| = \dfrac{|S|}{2} = \dfrac{9\,000\,000}{2} = 4\,500\,000$$.

Write the answer in the required form $$m \cdot n \cdot 10^{\,n}$$ with $$m,n \in \{1,2,\ldots ,9\}$$.

Observe $$4\,500\,000 = 45 \times 10^{5}$$.

Factor the coefficient $$45$$ as the product of two single-digit numbers: $$45 = 9 \times 5$$.

Thus $$m = 9,\; n = 5$$ and the requested sum is

$$m + n = 9 + 5 = 14.$$

Final answer: $$14$$.

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 $$\boxed{13}$$.

Divisible by 2 and 3 means the number should also be divisible by 6 and it should not be divisible by 4 and 9.

Smallest 3-digit multiple of 6 = 102

Largest = 996

Number of multiples: $$\dfrac{\left(996-102\right)}{6}+1=150$$

Now, we need to subtract the numbers divisible by both 4 and 9 (i.e., multiples of 36)

Smallest = 108, Largest = 972

$$\dfrac{\left(972-108\right)}{36}+1=25$$

Final answer: 150 - 25 = 125

We need to find the number of sequences of ten terms, where each term is 0, 1, or 2, containing exactly five 1s and exactly three 2s.

Since the sequence has 10 terms with exactly five 1s and exactly three 2s, the remaining $$10 - 5 - 3 = 2$$ terms must be 0s.

We need to choose positions for the 1s, 2s, and 0s among the 10 positions. This is a multinomial coefficient:

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

Computing this: $$\frac{10!}{5! \cdot 3! \cdot 2!} = \frac{3628800}{120 \cdot 6 \cdot 2} = \frac{3628800}{1440} = 2520$$.

Hence, the correct answer is Option C.

We are given the binomial coefficients $$\binom{n}{r-1}=28\,,\quad \binom{n}{r}=56\,,\quad \binom{n}{r+1}=70$$ and the points $$A(4\cos t,4\sin t)\,,\;B(2\sin t,-2\cos t)\,,\;C(3r-n,\;r^2-n-1)\,. $$ The locus of the centroid of triangle ABC is$$(3x-1)^2+(3y)^2=\alpha$$ and we must find $$\alpha$$.

First, from the ratio $$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{56}{28}=2=\frac{n-r+1}{r}$$ we get $$n-r+1=2r\quad\Rightarrow\quad n=3r-1\,. $$ Next, from $$\frac{\binom{n}{r+1}}{\binom{n}{r}}=\frac{70}{56}=\frac{5}{4}=\frac{n-r}{r+1}$$ we obtain $$4(n-r)=5(r+1)\quad\Rightarrow\quad 4n=9r+5\,. $$ 

Substituting $$n=3r-1$$ into $$4n=9r+5$$ gives $$4(3r-1)=9r+5\;\Rightarrow\;12r-4=9r+5\;\Rightarrow\;3r=9\;\Rightarrow\;r=3\,,\;n=8\,. $$ 

With $$r=3$$ and $$n=8$$, point $$C$$ is $$C=(3r-n,\;r^2-n-1)=(9-8,\;9-8-1)=(1,0)\,. $$ 

The centroid $$G$$ of triangle $$ABC$$ has coordinates $$x_G=\frac{4\cos t+2\sin t+1}{3},\qquad y_G=\frac{4\sin t-2\cos t+0}{3}\,. $$ Hence $$3x_G-1=4\cos t+2\sin t,\qquad3y_G=4\sin t-2\cos t\,. $$

Therefore the locus satisfies $$(3x_G-1)^2+(3y_G)^2=(4\cos t+2\sin t)^2+(4\sin t-2\cos t)^2$$ $$=16\cos^2t+16\sin t\cos t+4\sin^2t+16\sin^2t-16\sin t\cos t+4\cos^2t$$ $$=16(\cos^2t+\sin^2t)+4(\sin^2t+\cos^2t)=16+4=20\,. $$

Hence $$\alpha=20$$, which corresponds to Option D.

We need to find the number of 7-digit numbers formed using only digits 1, 2, and 3 such that the sum of digits equals 11.

Let the seven digits be $$x_1, x_2, \ldots, x_7$$ where each $$x_i \in \{1, 2, 3\}$$ and $$x_1 + x_2 + \cdots + x_7 = 11$$.

Let $$y_i = x_i - 1$$, so $$y_i \in \{0, 1, 2\}$$. Then:

$$ \sum_{i=1}^{7} (y_i + 1) = 11 \implies \sum_{i=1}^{7} y_i = 4 $$

We need to count the number of solutions to $$y_1 + y_2 + \cdots + y_7 = 4$$ where $$0 \leq y_i \leq 2$$.

Without the upper bound constraint, the number of non-negative integer solutions is $$\binom{4 + 7 - 1}{7 - 1} = \binom{10}{6} = 210$$.

Now subtract cases where any $$y_i \geq 3$$. If $$y_i \geq 3$$, let $$z_i = y_i - 3 \geq 0$$. Then the sum becomes $$z_i + \sum_{j \neq i} y_j = 1$$. The number of non-negative solutions is $$\binom{1 + 6}{6} = \binom{7}{6} = 7$$.

There are $$\binom{7}{1} = 7$$ ways to choose which variable exceeds 2, giving $$7 \times 7 = 49$$ cases.

No two variables can simultaneously be $$\geq 3$$ since that would require a sum $$\geq 6 > 4$$. So higher-order inclusion-exclusion terms are zero.

$$ N = 210 - 49 = 161 $$

The correct answer is Option 4: 161.

We need to select 10 players from 7 batsmen and 6 bowlers, including a designated captain (batsman) and vice-captain (bowler), with at least 4 batsmen and 4 bowlers total.

Since the captain (1 batsman) and vice-captain (1 bowler) are already included, we need 8 more players from the remaining 6 batsmen and 5 bowlers, with at least 3 more batsmen and 3 more bowlers.

The possible distributions of (additional batsmen, additional bowlers) are:

Case 1: (3, 5): $$\binom{6}{3} \times \binom{5}{5} = 20 \times 1 = 20$$

Case 2: (4, 4): $$\binom{6}{4} \times \binom{5}{4} = 15 \times 5 = 75$$

Case 3: (5, 3): $$\binom{6}{5} \times \binom{5}{3} = 6 \times 10 = 60$$

Total number of selections = $$20 + 75 + 60 = 155$$.

Hence, the correct answer is Option B.

The word "KANPUR" has 6 distinct letters: A, K, N, P, R, U.

Arranging in alphabetical order: A, K, N, P, R, U.

Total permutations = 6! = 720.

Words starting with each letter = 5! = 120.

Words starting with A: positions 1-120

Words starting with K: positions 121-240

Words starting with N: positions 241-360

Words starting with P: positions 361-480

440th word starts with P. Position within P-words: 440 - 360 = 80.

After P, remaining letters: A, K, N, R, U (alphabetical order).

Words starting with PA: 4! = 24 (positions 1-24)

Words starting with PK: 4! = 24 (positions 25-48)

Words starting with PN: 4! = 24 (positions 49-72)

Words starting with PR: 4! = 24 (positions 73-96)

80th word starts with PR. Position within PR-words: 80 - 72 = 8.

After PR, remaining letters: A, K, N, U.

Words starting with PRA: 3! = 6 (positions 1-6)

Words starting with PRK: 3! = 6 (positions 7-12)

8th word starts with PRK. Position within PRK-words: 8 - 6 = 2.

After PRK, remaining letters: A, N, U.

Words starting with PRKA: 2! = 2 (positions 1-2)

2nd word starting with PRKA: remaining letters N, U arranged as UN.

PRKAUN (position 2 within PRKA-words)

Wait: 1st position is PRKANU, 2nd position is PRKAUN.

So the 440th word is PRKAUN.

The correct answer is Option 3: PRKAUN.

The underlying set is $$A=\{a,b,c,d,e,f\}$$ with $$|A|=6$$.

A relation on $$A$$ is a subset of $$A\times A$$. The required relations must satisfy two conditions:

1. Reflexive: for every $$x\in A$$, the ordered pair $$(x,x)$$ is in the relation.   Therefore the six diagonal pairs   $$\{(a,a),(b,b),(c,c),(d,d),(e,e),(f,f)\}$$   are compulsory.   So each admissible relation already contains $$6$$ pairs.

2. Symmetric: if $$(x,y)$$ is in the relation, then so is $$(y,x)$$.

The problem states that the relation must contain exactly $$10$$ ordered pairs. Since $$6$$ pairs are fixed by reflexivity, we still have to add $$10-6=4$$ more ordered pairs.

Because of symmetry, the non-diagonal pairs must be chosen in symmetric blocks: adding $$(x,y)$$ automatically forces us to add $$(y,x)$$. Consequently the extra $$4$$ pairs must come in $$\frac{4}{2}=2$$ symmetric blocks.

How many off-diagonal symmetric blocks exist? Each block corresponds to an unordered pair $$\{x,y\}$$ with $$x\neq y$$. The number of such unordered pairs in a 6-element set is $$\binom{6}{2}=15$$.

To create the relation we simply pick any $$2$$ of these $$15$$ unordered pairs; each chosen pair contributes its two ordered pairs and thus supplies the needed $$4$$ additional elements.

Hence the number of admissible relations is $$\binom{15}{2}=105$$.

Therefore, the set $$\mathcal{S}$$ contains $$\mathbf{105}$$ relations.

We have a total of 22 points:
  • the origin $$O$$,
  • $$12$$ points $$P_1, P_2,\,\ldots,\,P_{12}$$ on the line $$L_1$$ (slope $$2$$),
  • $$9$$ points $$Q_1, Q_2,\,\ldots,\,Q_{9}$$ on the line $$L_2$$ (slope $$\tfrac12$$).

A triangle can be formed by any three non-collinear points. Hence

Total number of triangles
$$=\;\bigl(\text{all possible triples of points}\bigr)\;-\;\bigl(\text{collinear triples}\bigr).$$

Step 1: Count all possible triples of points
There are $$22$$ points in all, so the number of ways to choose any three of them is
$$\binom{22}{3} \;=\; \frac{22 \times 21 \times 20}{6} \;=\; 1540.$$

Step 2: Subtract triples that are collinear

• Triples on line $$L_1$$: Along $$L_1$$ we have the origin $$O$$ plus the 12 points $$P_i$$, altogether $$13$$ collinear points. The number of triples drawn entirely from these $$13$$ points is
$$\binom{13}{3} \;=\; \frac{13 \times 12 \times 11}{6} \;=\; 286.$$

• Triples on line $$L_2$$: Along $$L_2$$ we have the origin $$O$$ plus the 9 points $$Q_j$$, altogether $$10$$ collinear points. The number of triples drawn entirely from these $$10$$ points is
$$\binom{10}{3} \;=\; \frac{10 \times 9 \times 8}{6} \;=\; 120.$$

Step 3: Adjust for any overlap
A triple can lie on both $$L_1$$ and $$L_2$$ only if all three of its points are common to the two lines. The only common point of $$L_1$$ and $$L_2$$ is the origin $$O$$, so no triple is counted in both collinear groups. Therefore, no further adjustment is required.

Step 4: Final count of triangles
$$\begin{aligned} \text{Number of triangles} &= 1540 \;-\; 286 \;-\; 120 \\ &= 1134. \end{aligned}$$

Hence, the total number of distinct triangles that can be formed is $$1134$$, which corresponds to Option B.

First, treat the three teams $$X$$ (size $$2$$), $$Y$$ (size $$3$$) and $$Z$$ (size $$4$$) as labelled; changing a student from one team to another gives a different arrangement.

Total arrangements without any restriction
Choose the two members of $$X$$, then the three members of $$Y$$; the remaining four automatically form $$Z$$:
$$T = \binom{9}{2}\,\binom{7}{3} = 36 \times 35 = 1260.$$

Introduce two “bad” events:
  $$A$$ : student $$s_1$$ is placed in team $$X$$ (not allowed).
  $$B$$ : student $$s_2$$ is placed in team $$Y$$ (not allowed).

We need the count of arrangements that avoid both $$A$$ and $$B$$. Use the inclusion-exclusion principle:

Number of valid arrangements $$= T - |A| - |B| + |A\cap B|.$$

Count |A| (arrangements with $$s_1$$ in $$X$$)
Fix $$s_1$$ in $$X$$, pick one more member for $$X$$ from the remaining $$8$$ students, then form $$Y$$ from the remaining $$7$$ students:
$$|A| = \binom{8}{1}\,\binom{7}{3} = 8 \times 35 = 280.$$

Count |B| (arrangements with $$s_2$$ in $$Y$$)
Fix $$s_2$$ in $$Y$$, pick two more members for $$Y$$ from the remaining $$8$$ students, then form $$X$$ from the remaining $$6$$ students:
$$|B| = \binom{8}{2}\,\binom{6}{2} = 28 \times 15 = 420.$$

Count |A ∩ B| (both $$s_1$$ in $$X$$ and $$s_2$$ in $$Y$$)
Place $$s_1$$ in $$X$$ and choose one more member for $$X$$ from the $$7$$ students other than $$s_2$$: $$\binom{7}{1}=7$$.
Now $$s_2$$ is in $$Y$$; pick the remaining two members of $$Y$$ from the $$6$$ students still free: $$\binom{6}{2}=15$$.
$$|A\cap B| = 7 \times 15 = 105.$$

Apply inclusion-exclusion
$$\text{Valid arrangements}=1260 - 280 - 420 + 105 = 665.$$

Hence, the required number of ways to form the teams is 665.

We need to find the 50th word when all arrangements of the letters of BHBJO are written in dictionary order.

Letters of BHBJO: B, H, B, J, O — sorted in alphabetical order: B, B, H, J, O.

Total words = $$\frac{5!}{2!} = 60$$ (since B repeats twice).

Dictionary ordering — count words starting with each letter:

Words starting with B: Remaining letters are {B, H, J, O} (4 distinct). Number of arrangements = $$4! = 24$$. (Words 1-24)

Words starting with H: Remaining letters are {B, B, J, O}. Number of arrangements = $$\frac{4!}{2!} = 12$$. (Words 25-36)

Words starting with J: Remaining letters are {B, B, H, O}. Number of arrangements = $$\frac{4!}{2!} = 12$$. (Words 37-48)

Words starting with O: Remaining letters are {B, B, H, J}. Number of arrangements = $$\frac{4!}{2!} = 12$$. (Words 49-60)

The 50th word is the 2nd word among those starting with O.

Words starting with O, arranged in dictionary order:

Remaining letters: {B, B, H, J}, sorted: B, B, H, J.

Words starting with OB: Remaining {B, H, J} = $$3! = 6$$ words. (Words 49-54)

The 49th word is the first word starting with OB. Let's list:

- 49th: O B B H J

- 50th: O B B J H

So the 50th word is OBBJH.

The correct answer is Option (2): OBBJH.

$$\frac{C(n+1,r+1)}{C(n,r)} = \frac{55}{35} = \frac{11}{7}$$. $$\frac{n+1}{r+1} = 11/7 \Rightarrow 7n+7 = 11r+11 \Rightarrow 7n = 11r+4$$.

$$\frac{C(n,r)}{C(n-1,r-1)} = \frac{35}{21} = \frac{5}{3}$$. $$\frac{n}{r} = 5/3 \Rightarrow n = 5r/3$$.

From second: $$r = 3n/5$$. Sub in first: $$7n = 11(3n/5)+4 = 33n/5+4$$. $$35n-33n = 20$$. $$2n = 20$$, $$n = 10$$. $$r = 6$$.

$$2n+5r = 20+30 = 50$$.

The correct answer is Option (1): 50.

We have, $$4!=24$$, $$5!=120$$, and $$3!=6$$

Therefore,

$$\alpha = \dfrac{24!}{24^6}$$  and  $$\beta = \dfrac{120!}{120^{24}}$$

Or

$$\alpha = \dfrac{24!}{2^{18}\times 3^6}$$  and  $$\beta = \dfrac{120!}{2^{72}\times 3^{24}\times 5^{24}}$$

In order for $$\alpha$$ to be an integer, $$24!$$ must contain $$2^{18}$$ and $$3^6$$

The highest power of $$2$$ in $$24!$$ is $$\lfloor \dfrac{24}{2} \rfloor + \lfloor \dfrac{24}{4} \rfloor + \lfloor \dfrac{24}{8} \rfloor + \lfloor \dfrac{24}{16} \rfloor + \dots = 12+6+3+1= 22$$ which is larger than $$18$$

Similarly, the highest power of $$3$$ in $$24!$$ is larger than $$6$$.

Thus, $$\alpha \in \mathbb{N}$$

In $$\beta$$, in order for $$\beta$$ to be an integer, $$120!$$ must contain $$2^{72}$$, $$3^{24}$$, and $$5^{24}$$

The highest power of $$2$$ in $$120!$$ is $$60+30+15+7+3+1= 116$$, which is larger than $$72$$. The highest power of $$3$$ in $$120!$$ is $$40+13+4+1= 58$$, which is also larger than $$24$$. The highest power of $$5$$ in $$120!$$ is $$24+4= 28$$, which is larger than $$2$$.

Thus, $$\beta \in \mathbb{N}$$

Option C is the correct answer.

Since both the books and the shelves are identical, this problem reduces to finding the number of partitions of 8 into at most 4 parts.

A partition of 8 into at most 4 parts means writing 8 as an unordered sum of at most 4 positive integers (where empty shelves correspond to parts of value 0).

Listing all partitions of 8 into at most 4 parts:

1 part:

8 = 8 → (8)

2 parts:

8 = 7+1, 6+2, 5+3, 4+4 → 4 partitions

3 parts:

8 = 6+1+1, 5+2+1, 4+3+1, 4+2+2, 3+3+2 → 5 partitions

4 parts:

8 = 5+1+1+1, 4+2+1+1, 3+3+1+1, 3+2+2+1, 2+2+2+2 → 5 partitions

Total = 1 + 4 + 5 + 5 = 15

Therefore, the correct answer is Option 4: 15.

We need to count triangles whose vertices are vertices of a regular octagon, but none of whose sides is a side of the octagon.

A regular octagon has 8 vertices. The total number of triangles formed by choosing any 3 vertices is given by $$\binom{8}{3} = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{6} = 56$$.

Using complementary counting, the desired number equals the total number of triangles minus those with at least one side of the octagon.

To count triangles with exactly one side of the octagon, note that the octagon has 8 sides. For each side, the third vertex must not be adjacent to either endpoint of that side, which excludes the two endpoints themselves and their two other neighbors, leaving $$8 - 4 = 4$$ valid choices for the third vertex. Hence there are $$8 \times 4 = 32$$ such triangles.

Triangles with exactly two sides of the octagon occur when the two sides are consecutive, sharing a common vertex, and each such pair of consecutive sides determines exactly one triangle formed by three consecutive vertices. Since there are 8 vertices, there are 8 such pairs, giving $$8$$ triangles.

It is impossible for a triangle to have three sides of the octagon, so that count is $$0$$.

Therefore, the total number of triangles with at least one side of the octagon is $$32 + 8 + 0 = 40$$, and the number of triangles with no side of the octagon is $$56 - 40 = 16$$.

The correct answer is Option 4: 16.

We need to find the number of ways to choose 5 alphabets from the word MATHEMATICS (not necessarily distinct).

The distinct letters in MATHEMATICS with their frequencies are:

M: 2, A: 2, T: 2, H: 1, E: 1, I: 1, C: 1, S: 1

So we have 8 distinct letters, with 3 letters (M, A, T) appearing twice and 5 letters (H, E, I, C, S) appearing once.

We need to choose 5 letters where the chosen set respects the available multiplicities. We consider cases based on the pattern of repetitions.

Case 1: All 5 distinct.

Choose 5 from 8 distinct letters: $$\binom{8}{5} = 56$$.

Case 2: One pair + 3 distinct.

Choose which letter is repeated (from M, A, T): $$\binom{3}{1} = 3$$ ways.

Choose remaining 3 distinct letters from the other 7: $$\binom{7}{3} = 35$$ ways.

Total: $$3 \times 35 = 105$$.

Case 3: Two pairs + 1 distinct.

Choose 2 letters to be repeated (from M, A, T): $$\binom{3}{2} = 3$$ ways.

Choose 1 distinct letter from the remaining 6: $$\binom{6}{1} = 6$$ ways.

Total: $$3 \times 6 = 18$$.

Case 4: Three or more of the same letter.

Not possible since no letter appears more than twice.

Grand total = $$56 + 105 + 18 = 179$$.

The correct answer is Option 1: 179.

NAGPUR has 6 distinct letters: A,G,N,P,R,U. Alphabetical: A,G,N,P,R,U.

Words starting with A: 5! = 120. Starting with G: 120. Total = 240.

Words starting with N: 241-360. First 2 letters NA: 4! = 24 (241-264). NG: 24 (265-288). NP: 24 (289-312).

Word 313-336 start with NR. 313: NRAGPU, 314: NRAGUP, 315: NRAPGU.

The correct answer is Option (3): NRAPGU.

This is the problem of partitioning a set of 5 distinct elements into $$1, 2, 3,$$ or $$4$$ non-empty subsets (since offices are indistinguishable, we use Stirling numbers of the second kind, $$S(n, k)$$).

• 1 office used: $$S(5,1) = 1$$

• 2 offices used: $$S(5,2) = \frac{2^5 - 2}{2} = 15$$

• 3 offices used: $$S(5,3) = 25$$

• 4 offices used: $$S(5,4) = 10$$

Total ways $$n = 1 + 15 + 25 + 10 = \mathbf{51}$$.

We need to find the number of integers in $$[100, 700]$$ that are neither multiples of 3 nor multiples of 4. From 100 to 700 inclusive there are $$700 - 100 + 1 = 601$$ integers. To determine how many are multiples of 3, note that the number of multiples of 3 up to n is $$\lfloor n/3 \rfloor$$, so in this range one finds $$\lfloor 700/3 \rfloor - \lfloor 99/3 \rfloor = 233 - 33 = 200$$ multiples of 3. Similarly, the count of multiples of 4 is given by $$\lfloor 700/4 \rfloor - \lfloor 99/4 \rfloor = 175 - 24 = 151$$. Since LCM(3, 4) = 12, the integers divisible by both 3 and 4 correspond to multiples of 12, namely $$\lfloor 700/12 \rfloor - \lfloor 99/12 \rfloor = 58 - 8 = 50$$.

Applying the inclusion-exclusion principle, the total number of integers that are multiples of 3 or 4 is $$|A \cup B| = |A| + |B| - |A \cap B| = 200 + 151 - 50 = 301$$. Therefore, the complementary set of integers that are neither multiples of 3 nor multiples of 4 has size $$601 - 301 = 300$$.

The correct answer is Option 3: 300.

21 apples, 3 children, each ≥ 2. 

Give 2 each first: remaining 15 among 3 with each ≥ 0.

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

Total points: 5 on AB, 6 on BC, 7 on CA = 18 points total.

Total triangles from 18 points = C(18,3) = 816

Subtract collinear cases (3 points on same side don't form a triangle):

C(5,3) + C(6,3) + C(7,3) = 10 + 20 + 35 = 65

Number of valid triangles = 816 - 65 = 751

The correct answer is Option 3: 751.

The relation in the question is $$^{n-1}C_r = (k^{2}-8)\;{}^{n}C_{\,r+1}$$, where $$n$$ and $$r$$ are integers satisfying $$n \ge 1$$ and $$0 \le r \le n-1$$.

First write both binomial coefficients in factorial form.
$$^{n-1}C_r = \dfrac{(n-1)!}{r!\,(n-1-r)!}$$
$${}^{n}C_{\,r+1} = \dfrac{n!}{(r+1)!\,(n-r-1)!}$$

Compute their ratio.
$$\dfrac{^{\,n-1}C_r}{{}^{\,n}C_{\,r+1}} = \dfrac{\dfrac{(n-1)!}{r!\,(n-1-r)!}} {\dfrac{n!}{(r+1)!\,(n-r-1)!}} = \dfrac{(n-1)!}{r!\,(n-1-r)!}\; \dfrac{(r+1)!\,(n-r-1)!}{n!}$$

Simplify step by step.
• Because $$(r+1)! = (r+1)\,r!$$, the factor $$r!$$ cancels leaving $$r+1$$.
• Because $$(n-1-r)! = (n-r-1)!$$, these factors cancel completely.
• Use $$n! = n\,(n-1)!$$ to cancel $$(n-1)!$$.
After cancellations we obtain
$$\dfrac{^{\,n-1}C_r}{{}^{\,n}C_{\,r+1}} = \dfrac{r+1}{n}$$ $$-(1)$$

Return to the original equation and divide both sides by $${}^{\,n}C_{\,r+1}$$:
$$\dfrac{^{\,n-1}C_r}{{}^{\,n}C_{\,r+1}} = k^{2}-8$$
Using $$(1)$$ gives
$$k^{2}-8 = \dfrac{r+1}{n}$$ $$-(2)$$

Now analyse the fraction on the right side.
Because $$0 \le r \le n-1$$, we have $$1 \le r+1 \le n$$.
Hence
$$0 \lt \dfrac{r+1}{n} \le 1$$ $$-(3)$$

Combine $$(2)$$ and $$(3)$$ to bound $$k$$.
From $$(2)$$ and the lower bound in $$(3)$$:
$$k^{2}-8 \gt 0 \;\;\Longrightarrow\;\; k^{2} \gt 8 \;\;\Longrightarrow\;\; k \gt 2\sqrt{2}$$

From $$(2)$$ and the upper bound in $$(3)$$:
$$k^{2}-8 \le 1 \;\;\Longrightarrow\;\; k^{2} \le 9 \;\;\Longrightarrow\;\; k \le 3$$

Together,
$$2\sqrt{2} \lt k \le 3$$

This interval matches Option A.

Therefore, the given equation holds if and only if
Case Answer: Option A, $$2\sqrt{2} \lt k \le 3$$.

First fix the symbol conventions.
Combinations : $${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$$
Permutations : $$^{n}P_{r} = \frac{n!}{(n-r)!} = n(n-1)\dots (n-r+1)$$

Case 1 : Finding $$m$$ from the inequality

The given condition is
$${}^{6}C_{m} + 2\;{}^{6}C_{m+1} + {}^{6}C_{m+2} \gt {}^{8}C_{3}$$

Compute the right-hand side:
$${}^{8}C_{3} = \frac{8\cdot7\cdot6}{3\cdot2\cdot1}=56$$

The terms $${}^{6}C_{r}$$ are (for $$r=0$$ to $$6$$):
$$1,\;6,\;15,\;20,\;15,\;6,\;1$$

Because $$m+2 \le 6$$, the feasible values are $$m = 0,1,2,3,4$$. Evaluate the left-hand side for each:

$$\begin{aligned} m=0:&\;1 + 2(6) + 15 = 28\\ m=1:&\;6 + 2(15) + 20 = 56\\ m=2:&\;15 + 2(20) + 15 = 70\\ m=3:&\;20 + 2(15) + 6 = 56\\ m=4:&\;15 + 2(6) + 1 = 28 \end{aligned}$$

The inequality $$\gt56$$ is satisfied only for $$m = 2$$.

Case 2 : Finding $$n$$ from the permutation ratio

The second relation is
$$^{\,n-1}P_{3} : {}^{\,n}P_{4} = 1 : 8$$

Write both permutations explicitly:

$$^{\,n-1}P_{3} = (n-1)(n-2)(n-3)$$
$$^{\,n}P_{4} = n(n-1)(n-2)(n-3)$$

Therefore $$\frac{^{\,n-1}P_{3}}{^{\,n}P_{4}} = \frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)} = \frac{1}{n}$$

Given that this ratio equals $$\frac{1}{8}$$, we get $$n = 8$$.

Case 3 : Evaluating $$^{\,n}P_{m+1} + {}^{\,n+1}C_{m}$$

With $$n = 8$$ and $$m = 2$$, we need $$^{\,8}P_{3} + {}^{\,9}C_{2}$$.

Compute each term:

$$^{\,8}P_{3} = 8\cdot7\cdot6 = 336$$
$${}^{\,9}C_{2} = \frac{9\cdot8}{2\cdot1} = 36$$

Add them:
$$336 + 36 = 372$$

Hence $$^{\,n}P_{m+1} + {}^{\,n+1}C_{m} = 372$$.

Option D is correct.

The set $$S = \{2, 4, 8, 16, 32, 64, 128, 256, 512\}$$ has 9 elements ($$2^1$$ to $$2^9$$).

We need to partition $$S$$ into 3 sets $$A, B, C$$ with equal number of elements, so each set has 3 elements.

Count total ways to partition 9 elements into 3 groups of 3.

The number of ways to choose 3 elements for set A from 9: $$\binom{9}{3}$$

Then choose 3 from remaining 6 for set B: $$\binom{6}{3}$$

The remaining 3 go to set C: $$\binom{3}{3}$$

Since the sets A, B, C are distinguishable (they are labeled), we don't divide by $$3!$$.

$$ \text{Total partitions} = \binom{9}{3} \times \binom{6}{3} \times \binom{3}{3} = 84 \times 20 \times 1 = 1680 $$

The correct answer is Option (1): 1680.

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.

Find the serial number of the word TOUGH when all permutations of OUGHT are arranged in dictionary order.

Listing the letters in alphabetical order gives G, H, O, T, U (positions 1, 2, 3, 4, 5).

Words starting with G: $$4! = 24$$ (positions 1-24), words starting with H: $$4! = 24$$ (positions 25-48), words starting with O: $$4! = 24$$ (positions 49-72). Words starting with T come next (positions 73 onwards).

With remaining letters G, H, O, U, words starting with TG: $$3! = 6$$ (positions 73-78), words starting with TH: $$3! = 6$$ (positions 79-84). Words starting with TO come next (position 85 onwards).

With remaining letters G, H, U, words starting with TOG: $$2! = 2$$ (positions 85-86), words starting with TOH: $$2! = 2$$ (positions 87-88). Words starting with TOU come next (position 89).

Among words starting with TOU, the remaining letters are G and H. Since G comes before H, TOUGH is the first in this list, corresponding to position 89.

The correct answer is Option A: $$89$$.

We want the number of ways to select $$a \in \{2, 4, 6, \ldots, 100\}$$ and $$b \in \{1, 3, 5, \ldots, 99\}$$ such that $$a + b$$ leaves remainder 2 when divided by 23.

Since $$a$$ is even and $$b$$ is odd, the sum $$a + b$$ is always odd. We require $$a + b \equiv 2 \pmod{23}$$. The smallest possible sum is $$2 + 1 = 3$$ and the largest is $$100 + 99 = 199$$. The odd numbers in this range that are congruent to 2 modulo 23 are $$25, 71, 117, 163$$.

For $$a + b = 25$$, we have $$a \in \{2, 4, \ldots, 24\}$$ and $$b = 25 - a$$. Each even $$a$$ from 2 to 24 produces an odd $$b$$ between 23 and 1, giving 12 pairs.

For $$a + b = 71$$, we have $$a \in \{2, 4, \ldots, 70\}$$ and $$b = 71 - a$$. There are 35 even values of $$a$$ from 2 to 70, so this yields 35 pairs.

For $$a + b = 117$$, the condition $$b = 117 - a \le 99$$ implies $$a \ge 18$$, and $$b \ge 1$$ implies $$a \le 116$$. Since $$a$$ must be even and at most 100, it runs through $$\{18, 20, \ldots, 100\}$$, which gives 42 pairs.

For $$a + b = 163$$, the inequality $$b = 163 - a \le 99$$ gives $$a \ge 64$$, and $$b \ge 1$$ gives $$a \le 162$$. Among the even values up to 100, this means $$a \in \{64, 66, \ldots, 100\}$$, yielding 19 pairs.

Adding these counts gives $$12 + 35 + 42 + 19 = 108$$.

The answer is Option C: 108.

We need to find the position of the word PUBLIC when all permutations of its letters are arranged in dictionary (alphabetical) order.

The letters of PUBLIC in alphabetical order are: B, C, I, L, P, U.

All 6 letters are distinct, so total arrangements = $$6! = 720$$.

To find the rank of P-U-B-L-I-C:

Position 1: P

Letters before P in the sorted list: B, C, I, L (4 letters).

Words starting with each = $$5! = 120$$.

Count = $$4 \times 120 = 480$$.

Position 2: U (remaining: B, C, I, L, U)

Letters before U: B, C, I, L (4 letters).

Words for each = $$4! = 24$$.

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

Position 3: B (remaining: B, C, I, L)

Letters before B: none. Count = $$0$$.

Position 4: L (remaining: C, I, L)

Letters before L: C, I (2 letters).

Count = $$2 \times 2! = 4$$.

Position 5: I (remaining: C, I)

Letters before I: C (1 letter). Count = $$1 \times 1! = 1$$.

Position 6: C

Count = $$0$$.

Rank = $$480 + 96 + 0 + 4 + 1 + 0 + 1 = 582$$.

The correct answer is Option D: 582.

We need to find the number of ways to transport 8 persons in 3 cars (of different makes), where each car can hold at most 3 persons.

Since each car holds at most 3 persons and the total is 8, the only valid partition is: 3 + 3 + 2 = 8. No other partition works (e.g., 4+3+1 would require a car with 4, exceeding the limit).

Since the three cars are distinguishable (different makes), we need to decide which car gets 2 persons and which two get 3 persons.

Method: First choose which car carries only 2 persons: there are $$\binom{3}{1} = 3$$ ways.

Then, choose 2 persons for that car: $$\binom{8}{2}$$ ways.

Then, choose 3 persons from the remaining 6 for the next car: $$\binom{6}{3}$$ ways.

The last 3 persons go to the remaining car: $$\binom{3}{3} = 1$$ way.

$$\text{Total} = 3 \times \binom{8}{2} \times \binom{6}{3} \times \binom{3}{3}$$

$$= 3 \times 28 \times 20 \times 1 = 1680$$

Alternative calculation: We can also compute this as $$\frac{8!}{3!3!2!} \times \frac{3!}{2!}$$, where $$\frac{8!}{3!3!2!} = 560$$ counts the ways to partition 8 into groups of (3,3,2), and $$\frac{3!}{2!} = 3$$ assigns distinguishable cars to these groups (dividing by $$2!$$ because the two groups of 3 are interchangeable in the partition, but then multiplied by $$3!$$ for assigning to distinct cars).

$$560 \times 3 = 1680$$

The correct answer is Option 3: 1680.

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}, or they could both be E (using both E’s), which gives 1 way.

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. If the multiset contains two E’s and two distinct consonants, there are $$\dfrac{4!}{2!} = 12$$ arrangements.

Combining these cases leads to the total count:

The answer is $$504$$.

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