Permutation Combination JEE Main previous year questions with solutions

5 solved JEE Main questions on Permutation Combination, free to read — no sign-in needed. The full chapter has 220 questions; sign in to attempt the remaining 215 in the exam simulator.

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1. Q1JEE Main 2026 (05 Apr, Shift 1)Arrangement under Constraint
   Two players $A$ and $B$ play a series of games of badminton. The player, who wins $5$ games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player $A$ wins the series is __________.
   Show answer & solution

   Answer: 126

   Player $A$ wins the series if $A$ wins $5$ games before $B$ wins $5$ games. The series can last for a minimum of $5$ games and a maximum of $9$ games. If player $A$ wins the series in exactly $n$ games, then $A$ must win the $n$-th game, and $A$ must win exactly $4$ out of the first $n-1$ games. The number of ways this can happen is given by $^{n-1}C_{4}$. The possible values for $n$ are $5, 6, 7, 8,$ and $9$. Total number of ways for $A$ to win the series is the sum of the number of ways $A$ can win in $5, 6, 7, 8,$ or $9$ games: Total ways $= ^{4}C_{4} + ^{5}C_{4} + ^{6}C_{4} + ^{7}C_{4} + ^{8}C_{4}$ Using the property $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$, we can simplify the sum: $^{4}C_{4} = ^{5}C_{5}$ $^{5}C_{5} + ^{5}C_{4} = ^{6}C_{5}$ $^{6}C_{5} + ^{6}C_{4} = ^{7}C_{5}$ $^{7}C_{5} + ^{7}C_{4} = ^{8}C_{5}$ $^{8}C_{5} + ^{8}C_{4} = ^{9}C_{5}$ Therefore, the total number of ways is: $^{9}C_{5} = \dfrac{9!}{5!4!} = \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$ Answer: $126$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Combination
   There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is
   1. A.230
   2. B.220
   3. C.200
   4. D.210
   Show answer & solution

   Answer: (D)

   ${ }^{12} \mathrm{C}_3-{ }^5 \mathrm{C}_3=210$
3. Q3JEE Main 2024 (01 Feb, Shift 1)Division and Distribution of Distinct Items
   If $n$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $n$ is equal to:
   1. A.$47$
   2. B.$53$
   3. C.$51$
   4. D.$43$
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   Answer: (C)

   Given, Rooms are identical, So, total ways to distribute $5$ employees into $4$ rooms are: $5,0,0,0\Rightarrow 1$ way $4,1,0,0\Rightarrow \dfrac{5!}{4!}=5$ ways $3,2,0,0\Rightarrow \dfrac{5!}{3!2!}=10$ ways $2,2,0,1\Rightarrow \dfrac{5!}{2!2!2!}=15$ ways $2,1,1,1\Rightarrow \dfrac{5!}{2!{\left(1!\right)}^{3}3!}=10$ ways $3,1,1,0\Rightarrow \dfrac{5!}{3!2!}=10$ ways Total $\Rightarrow 1+5+10+15+10+10=51$ ways
4. Q4JEE Main 2023 (08 Apr, Shift 1)Circular Permutation
   The number of ways, in which $5$ girls and $7$ boys can be seated at a round table so that no two girls sit together is
   1. A.$720$
   2. B.$126(5!{)}^{2}$
   3. C.$7{\left(360\right)}^{2}$
   4. D.$7{\left(720\right)}^{2}$
   Show answer & solution

   Answer: (B)

   Given, $7$ boys and $5$ girls are to be seated around a circular such that no two girls to be seated together, Now we know that $n$ objects can be arranged in a circle in $(n-1)!$ ways. Let us first arrange $7$ boys in circular arrangement in $\left(7-1\right)!$ ways. Now there will be $7$ gaps. So let us select any $5$ gaps out of $7$ gaps and arrange $5$ girls in the chosen gaps. This can be done in $C57\times 5!$ ways. Hence, required arrangements are $6!\times C57\times 5!$ $=6\times 5!\times \dfrac{7\times 6}{2}\times 5!$ $=126{\left(5!\right)}^{2}$. Therefore, required arrangements are $126{\left(5!\right)}^{2}$
5. Q5JEE Main 2022 (25 Jun, Shift 1)Division of Identical items
   The number of $3$-digit odd numbers, whose sum of digits is a multiple of $7$, is _____.
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   Answer: 63

   Let the three-digit odd number be $xyz$ Now $z=1/3/5/7/9$ $x+y+z=7/14/21$ [for sum of digit to be multiple of $7$] $x+y=6/4/2/13/11/9/7/5/20/18/16/14/12$ When $x+y=6\Rightarrow \left(1,5\right),\left(2,4\right),\left(3,3\right),\left(4,2\right),\left(5,1\right),\left(6,0\right)$ Total possibilities$=6$ When $x+y=4\Rightarrow \left(1,3\right),\left(2,2\right),\left(3,1\right),\left(4,0\right)$ Total possibilities$=4$ When $x+y=2\Rightarrow \left(1,1\right),\left(2,0\right)$ Total possibilities$=2$ When $x+y=13\Rightarrow \left(4,9\right),\left(5,8\right),\left(6,7\right),\left(7,6\right),\left(8,5\right),\left(9,4\right)$ Total possibilities$=6$ When $x+y=11\Rightarrow \left(2,9\right),\left(3,8\right),\left(4,7\right),\left(5,6\right),\left(6,5\right),\left(6,5\right),\left(7,4\right),\left(8,3\right),\left(9,2\right)$ Total possibilities$=8$ When $x+y=9\Rightarrow \left(1,8\right),\left(2,7\right),\left(3,8\right),\left(4,5\right),\left(5,4\right),\ldots .\left(8,1\right),\left(9,0\right)$ Total possibilities$=9$ When $x+y=7\Rightarrow \left(1,8\right),\left(2,5\right),\left(3,4\right),\ldots .\left(8,1\right),\left(7,0\right)$ Total possibilities$=7$ When $x+y=5\Rightarrow \left(1,4\right),\left(2,3\right),\left(3,2\right),\left(4,1\right),\left(5,0\right)$ Total possibilities$=5$ When $x+y=20$ No possibilities When $x+y=18\Rightarrow \left(9,9\right)$ Total possibilities$=1$ When $x+y=16\Rightarrow \left(7,9\right),\left(8,8\right),\left(9,7\right)$ Total possibilities$=3$ When $x+y=14\Rightarrow \left(5,9\right),\left(6,8\right),\left(7,7\right),\left(8,6\right),\left(9,5\right)$ Total possibilities$=5$ When $x+y=12\Rightarrow \left(3,9\right),\left(4,8\right),\left(5,7\right),\left(6,6\right)\ldots .\left(9,3\right)$ Total possibilities$=7$ Hence, the total number of three digit odd numbers whose sum of digit is divisible by $7$ is $63$