Permutation Combination JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Permutation Combination, free to read — no sign-in needed. The full chapter has 30 questions; sign in to attempt the remaining 25 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   A bookshelf contains $6$ distinct books of Mathematics and $5$ distinct books of Physics. From these $11$ books, $6$ books are chosen at random. Let $X$ be the absolute value of the difference between the number of Mathematics books chosen and the number of Physics books chosen. If $\alpha$ is the mean of the random variable $X$, then the value of $77\alpha$ is ___________.
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   Answer: 100

   Total number of ways to choose $6$ books from $11$ books is $^{11}C_{6} = 462$. Let $M$ and $P$ be the number of Mathematics and Physics books chosen respectively. Since $6$ books are chosen, $M + P = 6$. The random variable $X$ is given by $X = |M - P| = |2M - 6|$. The possible selections of $(M, P)$, the corresponding values of $X$, and the number of ways are as follows: For $M=1, P=5$: $X = |1 - 5| = 4$, Number of ways $= ^{6}C_{1} \times ^{5}C_{5} = 6 \times 1 = 6$ For $M=2, P=4$: $X = |2 - 4| = 2$, Number of ways $= ^{6}C_{2} \times ^{5}C_{4} = 15 \times 5 = 75$ For $M=3, P=3$: $X = |3 - 3| = 0$, Number of ways $= ^{6}C_{3} \times ^{5}C_{3} = 20 \times 10 = 200$ For $M=4, P=2$: $X = |4 - 2| = 2$, Number of ways $= ^{6}C_{4} \times ^{5}C_{2} = 15 \times 10 = 150$ For $M=5, P=1$: $X = |5 - 1| = 4$, Number of ways $= ^{6}C_{5} \times ^{5}C_{1} = 6 \times 5 = 30$ For $M=6, P=0$: $X = |6 - 0| = 6$, Number of ways $= ^{6}C_{6} \times ^{5}C_{0} = 1 \times 1 = 1$ The mean $\alpha$ of the random variable $X$ is given by: $\alpha = \dfrac{\sum X_i f_i}{\sum f_i}$ $\alpha = \dfrac{4(6) + 2(75) + 0(200) + 2(150) + 4(30) + 6(1)}{462}$ $\alpha = \dfrac{24 + 150 + 0 + 300 + 120 + 6}{462}$ $\alpha = \dfrac{600}{462} = \dfrac{100}{77}$ Therefore, the value of $77\alpha$ is: $77\alpha = 77 \times \dfrac{100}{77} = 100$ Answer: $100$
2. Q2JEE Advanced Adv 2026 (Paper 1)
   The number of ways to distribute $10$ identical red pens and $14$ identical blue pens among four persons such that each person gets $6$ pens, is _____________.
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   Answer: 206

   Let $r_i$ and $b_i$ be the number of red and blue pens given to the $i$-th person, where $i = 1, 2, 3, 4$. Given that each person receives exactly $6$ pens, we have: $r_i + b_i = 6 \Rightarrow b_i = 6 - r_i$ Since the number of blue pens must be non-negative ($b_i \ge 0$), we get $r_i \le 6$. Also, $r_i \ge 0$. The total number of red pens is $10$, so: $r_1 + r_2 + r_3 + r_4 = 10$ The number of ways to distribute the pens is the number of non-negative integer solutions to this equation subject to $0 \le r_i \le 6$. This is equivalent to finding the coefficient of $x^{10}$ in the expansion of $(1 + x + x^2 + \dots + x^6)^4$. $(1 + x + x^2 + \dots + x^6)^4 = \left( \dfrac{1 - x^7}{1 - x} \right)^4 = (1 - x^7)^4 (1 - x)^{-4}$ Expanding both terms: $(1 - x^7)^4 = 1 - 4x^7 + 6x^{14} - \dots$ $(1 - x)^{-4} = \sum_{k=0}^{\infty} {}^{k+3}C_{3} x^k$ We need the coefficient of $x^{10}$ in the product $(1 - 4x^7 + \dots) \sum_{k=0}^{\infty} {}^{k+3}C_{3} x^k$. Coefficient of $x^{10} = 1 \times (\text{coefficient of } x^{10} \text{ in } (1-x)^{-4}) - 4 \times (\text{coefficient of } x^3 \text{ in } (1-x)^{-4})$ $= {}^{10+3}C_{3} - 4 \times {}^{3+3}C_{3}$ $= {}^{13}C_{3} - 4 \times {}^{6}C_{3}$ $= \dfrac{13 \times 12 \times 11}{3 \times 2 \times 1} - 4 \times \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1}$ $= 286 - 4 \times 20$ $= 286 - 80 = 206$ Answer: $206$
3. Q3JEE Advanced Adv 2025 (Paper 1)
   Let $S$ be the set of all seven-digit numbers that can be formed using the digits 0,1 and 2 . For example, 2210222 is in $S$, but 0210222 is NOT in $S$. Then the number of elements $x$ in $S$ such that at least one the digits 0 and 1 appears exactly twice in $x$, is equal to ______.
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   Answer: 762

   Let $\mathrm{A} \rightarrow$ " 0 " appear exactly twice. and $\mathrm{B} \rightarrow " 1 "$ appear exactly twice. $\therefore \mathrm{A} \cap \mathrm{B} \rightarrow$ " 0 " and " 1 " both appears exactly twice. $\mathrm{n}(\mathrm{A})=-\underbrace{-------}$ $={ }_{\text {placing zero }}^6 \mathrm{C}_2(1)(2)^5=\frac{6 \times 5}{2} \times 2^5=480$ for $\mathrm{n}(\mathrm{B})$ C-I : $1$ at first place $\underline{1}\underbrace{-------}$ $\text { Number of ways }=\underset{\text { placing } 1}{} \mathrm{}^6{C}_1(1)(2)^5=192$ C-II : 2 at first place $\underline{2}\underbrace{-------}$ Number of ways $=\underset{\text { placing } 1}{{ }^6 \mathrm{C}_2(1)(2)^4}=\frac{6 \times 5}{2} \times 2^4=240$ $n(B)=240+192$ for $\mathrm{n}(\mathrm{A} \cap \mathrm{B})$ $\underline{}\underbrace{-------}$ for $\mathrm{n}(\mathrm{A} \cap \mathrm{B})=\underset{\text { placing zero }}{{ }^6 \mathrm{C}_2(1)} \times \underset{\text { placing } 1}{{ }^5 \mathrm{C}_2(1)} \times \underset{\text { 2at rest places } 1}{(1)} =\frac{6 \times 5}{2} \times \frac{5 \times 4}{2}=150$ $\begin{aligned} & \therefore \mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \\ & =480+(192+240)-150 \\ & =762\end{aligned}$
4. Q4JEE Advanced Adv 2024 (Paper 1)
   A group of 9 students, $\mathrm{s}_1, \mathrm{~s}_2, \ldots \ldots, \mathrm{s}_9$, is to be divided to from three teams $\mathrm{X}, \mathrm{Y}$, and $\mathrm{Z}$ of sizes 2,3 , and 4 , respectively. Suppose that $s_1$ cannot be selected for the team $\mathrm{X}$, and $\mathrm{s}_2$ cannot be selected for the team Y. Then the number of ways to from such teams, is
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   Answer: 665

   $\begin{array}{ccc}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 2 & 3 & 4 \end{array}$ C-i) when $x$ does not contain $\mathrm{S}_1$, but contains $\mathrm{S}_2$ $\underset{\text{(for }x)}{{ }^7 C_1} \times \underset{\text{(for }y,z)}{\frac{7!}{3!4!}}=245$ C-ii) When $\mathrm{x}$ does not contain $\mathrm{S}_1, \mathrm{~S}_2$ and $\mathrm{y}$ does not contain $\mathrm{S}_2$ i.e. $\underset{\text{(for }x)}{{ }^{7} \mathrm{C}_2} \times \underset{\text{(for }y,z)}{\frac{6!}{3!3!}}=420$ so total No. of ways $665$
5. Q5JEE Advanced Adv 2023 (Paper 2)
   Let $X$ be the set of all five digit numbers formed using $1,2,2,2,4,4,0$. For example, $22240$ is in $X$ while $02244$ and $44422$ are not in $X$. Suppose that each element of $X$ has an equal chance of being chosen. Let $p$ be the conditional probability that an element chosen at random is a multiple of $20$ given that it is a multiple of $5$. Then the value of $38p$ is equal to
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   Answer: 31

   First we will find the sample space in which the number of five-digit numbers are divisible by $5$, So, fixing zero at the last place we get, $----0_{⏟}$ Now in first four place following number can take place, $2224\rightarrow \dfrac{4!}{3!}=4\text{ways}$ $2244\rightarrow \dfrac{4!}{2!2!}=6\text{ways}$ $2221\rightarrow \dfrac{4!}{3!}=4\text{ways}$ $2241\rightarrow \dfrac{4!}{2!}=12\text{ways}$ $2441\rightarrow \dfrac{4!}{2!}=12\text{ways}$ So, total sample space will be $4+6+4+12+12=38$ Now finding the number of favourable outcomes, So, Number of five-digit numbers divisible by $5$ but 'not' by $20$ Now fixing $10$ in last two places, we get $---10_{⏟}$ So, the first three places can be occupied by, $222\rightarrow 1\text{ways}$ $224\rightarrow 3\text{ways}$ $244\rightarrow 3\text{ways}$ So, total number of numbers which are divisible by $5$ but not $20$ will be, $1+3+3=7$ So, favourable number of five-digit numbers that are divisible by $5$ and $20=38-7=31$ Hence, probability is given by, $p=\dfrac{31}{38}$ $\Rightarrow 38p=31$