Permutations & Combinations Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Permutations and Combinations with concise, step-by-step KaTeX solutions covering onto/one-one functions, arrangements with restrictions, integer solutions, and selection problems.
A curated set of Permutations & Combinations questions from JEE Main 2026, each worked out step by step so you can lock in the counting patterns the exam keeps repeating.
Solutions are AI-generated and pending review.
Solution
Total functions from a $4$-element set to a $3$-element set:
$$3^4 = 81.$$Number of onto (surjective) functions is counted by inclusion–exclusion:
$$\sum_{i=0}^{3}(-1)^i \binom{3}{i}(3-i)^4 = 3^4 - 3\cdot 2^4 + 3\cdot 1^4 = 81 - 48 + 3 = 36.$$Therefore the number of functions that are not onto:
$$81 - 36 = 45.$$Answer: B (45)
Solution
Total arrangements of $7$ distinct people:
$$7! = 5040.$$Count the arrangements where all 3 girls are together: treat the girls as one block, giving $5$ units to arrange, and the girls permute among themselves:
$$5! \times 3! = 120 \times 6 = 720.$$“Girls not all together” is the complement:
$$5040 - 720 = 4320.$$Answer: D (4320)
Solution
List the distinct letters of INCONSEQUENTIAL.
- Distinct vowels: $A, E, I, O, U$ — that is $5$ vowels.
- Distinct consonants: $C, L, N, Q, S, T$ — that is $6$ consonants.
Since no letter may be repeated, choose $2$ distinct vowels and $2$ distinct consonants, then arrange all $4$ chosen letters:
$$\binom{5}{2}\binom{6}{2}\,4! = 10 \times 15 \times 24 = 3600.$$Answer: D (3600)
Solution
Each of the $7$ positions gets one of the $5$ digits, and every digit must appear at least once. This is the number of onto functions from $7$ positions onto $5$ digits (all digits are nonzero, so no leading-zero issue arises).
By inclusion–exclusion:
$$\sum_{i=0}^{5}(-1)^i \binom{5}{i}(5-i)^7.$$Computing the terms:
$$5^7 - \binom{5}{1}4^7 + \binom{5}{2}3^7 - \binom{5}{3}2^7 + \binom{5}{4}1^7$$$$= 78125 - 5(16384) + 10(2187) - 10(128) + 5(1)$$$$= 78125 - 81920 + 21870 - 1280 + 5 = 16800.$$Answer: C (16800)
Solution
Fix $c$. Then $a + b = 22 - 2c$, and the number of non-negative integer pairs $(a, b)$ with a fixed sum $s = 22 - 2c$ is $s + 1$.
Since $a + b = 22 - 2c \ge 0$, we need $c = 0, 1, 2, \ldots, 11$. For each such $c$:
$$\text{count} = (22 - 2c) + 1 = 23 - 2c.$$Summing over $c = 0$ to $11$:
$$\sum_{c=0}^{11}(23 - 2c) = 23 + 21 + 19 + \cdots + 1 = \sum_{k=1,\,3,\,\ldots}^{23} k.$$This is the sum of the first $12$ odd numbers ($1, 3, \ldots, 23$):
$$12^2 = 144.$$Answer: C (144)
Solution
Triangles from the vertices of an $n$-gon are just choices of $3$ vertices:
$$P_n = \binom{n}{3}.$$Then, using $\binom{n+1}{3} - \binom{n}{3} = \binom{n}{2}$ (Pascal’s identity):
$$P_{n+1} - P_n = \binom{n}{2} = \frac{n(n-1)}{2} = 66.$$So $n(n-1) = 132 = 12 \times 11 \Rightarrow n = 12.$
The distinct prime divisors of $12 = 2^2 \cdot 3$ are $2$ and $3$, whose sum is:
$$2 + 3 = 5.$$Answer: C (5)
Solution
The lift does not stop at floors $1$ and $2$, so the available exit floors are $3, 4, \ldots, 10$, i.e. $8$ floors.
- Choose which 4 of the $9$ persons form the group that leaves together: $\binom{9}{4}$. The remaining $5$ are automatically the other group.
- Assign floors: the group of $4$ and the group of $5$ leave at two different floors. Pick an ordered pair of distinct floors (one for each distinguishable group): $8 \times 7 = 56$.
Answer: C (7056)
Solution
First handle the constraint $f(2) + f(3) = 5$ with $f(3) \le 4$ and $f(2), f(3)$ distinct (injective). The valid pairs $\big(f(3), f(2)\big)$ are:
$$(1,4),\ (2,3),\ (3,2),\ (4,1) \quad\text{— } 4 \text{ pairs.}$$For each fixed pair, two of the six output values are used up. Now count:
- $f(1) \ge 3$: choose $f(1)$ from the values in $\{3,4,5,6\}$ still available (i.e. not already taken by $f(2), f(3)$).
- The remaining $3$ values fill positions $4, 5, 6$ in $3! = 6$ ways.
Checking each pair, exactly $3$ of the values $\{3,4,5,6\}$ remain free for $f(1)$ (one of them is always consumed by $f(2)$ or $f(3)$). So each pair contributes:
$$3 \times 3! = 3 \times 6 = 18.$$Total over the $4$ pairs:
$$4 \times 18 = 72.$$Answer: 72
Solution
A wins the series on the $k$-th game, where the $k$-th game is A’s $5$-th win. So A must win exactly $4$ of the first $k-1$ games (and lose the rest), then win game $k$. Here $k$ ranges from $5$ to $9$.
The number of winning sequences is:
$$\sum_{k=5}^{9}\binom{k-1}{4} = \binom{4}{4} + \binom{5}{4} + \binom{6}{4} + \binom{7}{4} + \binom{8}{4}.$$$$= 1 + 5 + 15 + 35 + 70 = 126.$$Answer: 126
Solution
Let $b, y, r$ be the numbers of blue, yellow, red balls drawn, with
$$b + y + r = 8,\quad 2 \le b \le 5,\ 2 \le y \le 6,\ 2 \le r \le 4.$$Since the balls of a colour are distinguishable, the number of selections for a given $(b, y, r)$ is $\binom{5}{b}\binom{6}{y}\binom{4}{r}$.
Enumerating the valid triples (with $b + y + r = 8$):
$$\sum \binom{5}{b}\binom{6}{y}\binom{4}{r} = 4100.$$For example the leading contributions include $(2,4,2)$, $(2,2,4)$, $(4,2,2)$, $(3,3,2)$, etc.; summing all admissible cases gives $4100$.
Answer: A (4100)
Solution
Each of the $4$ distinct books goes into one of the $3$ distinct bags, and every bag must receive at least one book. This is the number of onto functions from $4$ books to $3$ bags.
By inclusion–exclusion:
$$\sum_{i=0}^{3}(-1)^i \binom{3}{i}(3-i)^4 = 3^4 - 3\cdot 2^4 + 3\cdot 1^4 = 81 - 48 + 3 = 36.$$Answer: B (36)