JEE Permutations & Combinations PYQs with Video Solutions PDF

The word “DISTRIBUTION” contains 12 letters in all.
The distinct letters and their multiplicities are: $$D_1, I_3, S_1, T_2, R_1, B_1, U_1, O_1, N_1.$$
Hence there are $$9$$ different letters, with only $$I$$ available thrice and $$T$$ available twice.

We have to form 4-letter arrangements (words) subject to these multiplicities.
Treat each possible repetition pattern separately and add the counts.

Case 1: All four letters different, pattern $$1+1+1+1$$.
Choose any $$4$$ different letters from $$9$$: $$\binom{9}{4}$$ ways.
Arrange them: $$4!$$ ways.
Count $$= \binom{9}{4}\,4! = 126 \times 24 = 3024.$$

Case 2: One letter repeated twice and two other distinct letters, pattern $$2+1+1$$.
• The repeated letter can only be $$I$$ or $$T$$ - 2 choices.
• Choose the other 2 distinct letters from the remaining 8 letters: $$\binom{8}{2}$$ ways.
• Arrange the multiset $$\{a,a,b,c\}$$: $$\dfrac{4!}{2!}=12$$ ways.
Count $$= 2 \times \binom{8}{2} \times 12 = 2 \times 28 \times 12 = 672.$$

Case 3: Two different letters each repeated twice, pattern $$2+2$$.
Possible only with $$I$$ and $$T$$, giving the multiset $$\{I,I,T,T\}$$ (1 way).
Arrange: $$\dfrac{4!}{2!\,2!}=6$$ ways.
Count $$=6.$$

Case 4: One letter repeated thrice and one other distinct letter, pattern $$3+1$$.
• The triple must be $$I$$ (only letter with 3 copies).
• Choose the remaining single letter from the 8 letters different from $$I$$: 8 ways.
• Arrange $$\{I,I,I,b\}$$: $$\dfrac{4!}{3!}=4$$ ways.
Count $$= 8 \times 4 = 32.$$

Case 5: All four letters identical, pattern $$4$$ - impossible because no letter appears four times.
Count $$=0.$$

Add all permissible cases: $$3024 + 672 + 6 + 32 = 3734.$$

Therefore, the required number of 4-letter words is 3734 .

The given ratio is

$${}^{\,n+1}C_{\,r+1} : {}^{\,n}C_{\,r} : {}^{\,n-1}C_{\,r-1}=55:35:21.$$

We compare the first two terms and the last two terms separately.

Case 1: $$\frac{{}^{\,n+1}C_{\,r+1}}{{}^{\,n}C_{\,r}}=\frac{55}{35}=\frac{11}{7}.$$

Using the factorial form, $$\frac{{}^{\,n+1}C_{\,r+1}}{{}^{\,n}C_{\,r}} =\frac{\dfrac{(n+1)!}{(r+1)!\,(n-r)!}}{\dfrac{n!}{r!\,(n-r)!}} =\frac{n+1}{r+1}.$$

Hence $$\frac{n+1}{r+1}=\frac{11}{7} \;\Longrightarrow\; 7(n+1)=11(r+1) \;\Longrightarrow\; 7n-11r=4.$$ Call this $$-(1).$$

Case 2: $$\frac{{}^{\,n}C_{\,r}}{{}^{\,n-1}C_{\,r-1}}=\frac{35}{21}=\frac{5}{3}.$$

Again, with factorials, $$\frac{{}^{\,n}C_{\,r}}{{}^{\,n-1}C_{\,r-1}} =\frac{\dfrac{n!}{r!\,(n-r)!}}{\dfrac{(n-1)!}{(r-1)!\,(n-r)!}} =\frac{n}{r}.$$

Therefore $$\frac{n}{r}=\frac{5}{3} \;\Longrightarrow\; 3n=5r.$$ Call this $$-(2).$$

Solve the simultaneous equations (1) and (2).

From (2): $$n=\frac{5r}{3}.$$

Substitute in (1):

$$7\left(\frac{5r}{3}\right)-11r=4 \;\Longrightarrow\; \frac{35r}{3}-11r=4 \;\Longrightarrow\; \frac{35r-33r}{3}=4 \;\Longrightarrow\; \frac{2r}{3}=4 \;\Longrightarrow\; r=6.$$

Then from (2): $$n=\frac{5(6)}{3}=10.$$

Finally, $$2n+5r=2(10)+5(6)=20+30=50.$$

Option A which is: $$50$$

Case 1: The number starts with 1 ($$\text{1\_ \_ \_ \_ \_}$$)

Any 6-digit number starting with $$1$$ is automatically less than $$331,214$$.

• Remaining digits to place: $$\{1, 2, 3, 3, 4\}$$
• We must choose an even ending digit to keep the number even:

Subcase 1a: Ends in 2

• The remaining middle 4 digits are $$\{1, 3, 3, 4\}$$.
• Since the digit $$3$$ is repeated twice, the number of distinct permutations is:$$\text{Ways} = \frac{4!}{2!} = \frac{24}{2} = 12$$

Subcase 1b: Ends in 4

• The remaining middle 4 digits are $$\{1, 2, 3, 3\}$$.
• Again, the digit $$3$$ is repeated twice:$$\text{Ways} = \frac{4!}{2!} = \frac{24}{2} = 12$$

$$\mathbf{\text{Total for Case 1}} = 12 + 12 = \mathbf{24\text{ ways}}$$

Case 2: The number starts with 2 ($$\text{2\_ \_ \_ \_ \_}$$)

Any 6-digit number starting with $$2$$ is also automatically less than $$331,214$$.

• Remaining digits to place: $$\{1, 1, 3, 3, 4\}$$
• Because the only $$2$$ has already been used at the front, the number must end in 4 to be even.

Subcase 2a: Ends in 4

• The remaining middle 4 digits are $$\{1, 1, 3, 3\}$$.
• Here, both $$1$$ and $$3$$ are repeated twice, so we divide by $$2! \cdot 2!$$:$$\text{Ways} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6$$

$$\mathbf{\text{Total for Case 2}} = \mathbf{6\text{ ways}}$$

Case 3: The number starts with 3 ($$\text{3\_ \_ \_ \_ \_}$$)

Since our target number is $331,214$, numbers starting with $3$ must be carefully filtered by their subsequent digits to ensure they remain strictly less than the target.

Subcase 3a: Starts with 31 ($$\text{31\_ \_ \_ \_}$$)

• Remaining digits to place: $\{1, 2, 3, 4\}$ (all unique)
• Ends in 2: Middle 3 digits are $\{1, 3, 4\} \implies 3! = 6$ ways.
• Ends in 4: Middle 3 digits are $\{1, 2, 3\} \implies 3! = 6$ ways.$$\text{Total for 31} = 6 + 6 = 12\text{ ways}$$

Subcase 3b: Starts with 32 ($$\text{32\_ \_ \_ \_}$$)

• Remaining digits to place: $\{1, 1, 3, 4\}$
• Since the $2$ is used, the number must end in 4 .
• Ends in 4: Middle 3 digits are $\{1, 1, 3\}$. Because $1$ repeats twice:$$\text{Ways} = \frac{3!}{2!} = 3\text{ ways}$$$$\text{Total for 32} = 3\text{ ways}$$

Subcase 3c: Starts with 33 ($$\text{33\_ \_ \_ \_}$$)

• Remaining digits to place: $$\{1, 1, 2, 4\}$$
• To be strictly less than $$33\mathbf{1},214$$, the third digit must be $$1$$. This locks our prefix to $$\text{331}$$.
• Remaining digits after $$\text{331}$$: $$\{1, 2, 4\}$$
  • Starts with 3311: The remaining digits are $$\{2, 4\}$$. This leaves only 2 permutations: $$331124$$ and $$331142$$. Both are even and valid $$\implies 2$$ ways.
  • Starts with 3312: The remaining digits are $$\{1, 4\}$$. To be even, it must end in $$4$$, giving the number $$331214$$. However, this is equal to our target, not strictly less than it $$\implies 0$$ ways.$$\text{Total for 33} = 2 + 0 = 2\text{ ways}$$

$$\mathbf{\text{Total for Case 3}} = 12 + 3 + 2 = \mathbf{17\text{ ways}}$$

Final Summary

Since these three primary cases are mutually exclusive, we find the total number of valid even configurations by summing them together:

$$\text{Total Even Numbers} = \text{Case 1} + \text{Case 2} + \text{Case 3}$$  $$\text{Total Even Numbers} = 24 + 6 + 17 = 47$$

There are exactly 47 distinct even numbers less than $$331,214$$ that can be formed.

The letters of $$\text{MATRIX}$$ in alphabetical order are: $$A,I,M,R,T,X$$

Each starting letter contributes $$5!=120$$ words.

Ranks $$1\text{ to }120$$ start with $$A,$$

ranks $$121\text{ to }240$$ start with $$I,$$ and

ranks $$241\text{ to }360$$ start with $$M.$$

Therefore,

$$W_1$$ starts with $$M.$$

Now fix $$M$$ and examine the second letter.

Each choice contributes $$4!=24$$ words.

$$MA:\ 241\text{ to }264$$

$$MI:\ 265\text{ to }288$$

$$MR:\ 289\text{ to }312$$

Hence,

$$313\text{ onwards}$$ begin with $$MT.$$

Now arrange the remaining letters $$A,I,R,X$$ in dictionary order:

1. $$MTAIRX\quad (313)$$

2. $$MTAIXR\quad (314)$$

Therefore,

$$W_1=\text{MTAIXR}$$

Arrange the first three letters $$M,T,A$$ alphabetically: $$A,M,T$$

Keeping the last three letters unchanged, $$W_2=\text{AMTIXR}$$

Now find the rank of $$\text{AMTIXR}.$$

Words beginning with $$A$$ come first.

Fix $$A.$$

Now consider the second letter: $$AI:\ 4!=24$$

Hence,

$$25^\text{th}$$ word onwards begin with $$AM.$$

Fix $$AM.$$

Now consider the third letter:

$$AMI:\ 3!=6$$

$$AMR:\ 3!=6$$

Therefore,

$$37^\text{th}$$ word onwards begin with $$AMT.$$

Fix $$AMT.$$

Remaining letters are $$I,R,X.$$

Their dictionary order starts as:

1. $$AMTIRX\quad (37)$$

2. $$AMTIXR\quad (38)$$

Hence, the rank of $$W_2$$ is

$$\boxed{38}$$

The units digits of $$2^n$$ repeat as $$2,4,8,6$$ with period $$4.$$

The units digits of $$8^n$$ repeat as $$8,4,2,6$$ with period $$4.$$

Adding corresponding units digits:

$$n\equiv1\pmod4:\quad 2+8=10$$

$$n\equiv2\pmod4:\quad 4+4=8$$

$$n\equiv3\pmod4:\quad 8+2=10$$

$$n\equiv0\pmod4:\quad 6+6=12$$

Therefore, $$2^n+8^n$$ is divisible by $$10$$ when $$n\equiv1\pmod4$$ or $$n\equiv3\pmod4.$$

There are $$9999-1000+1=9000$$ four-digit numbers.

Exactly half satisfy $$n\equiv1\pmod4$$ or $$n\equiv3\pmod4.$$

Hence the required count is $$\frac{9000}{2}=4500.$$