Permutations and Combinations: JEE Main PYQ, 2020–2025

Answer: (B)

The number of ways to choose 2 objects from n is C(n, 2) = n(n − 1)/2. Setting n(n − 1)/2 = 45 gives n(n − 1) = 90, so n² − n − 90 = 0, which factors as (n − 10)(n + 9) = 0. Since n must be a positive integer, n = 10. Check: C(10, 2) = (10 × 9)/2 = 45.

Answer: (A)

Split by the last digit. Case 1: the last digit is 0. The other three places are filled from the remaining 9 digits in P(9,3) = 9×8×7 = 504 ways. Case 2: the last digit is 2, 4, 6 or 8 (4 ways). The first digit can be any of the remaining digits except 0, so 8 ways. The two middle places are filled from the 8 digits left in 8×7 = 56 ways. This gives 4 × 8 × 56 = 1792. Total = 504 + 1792 = 2296.

Answer: (A)

Let the chosen numbers be a < b < c with b − a ≥ 2 and c − b ≥ 2. Put a' = a, b' = b − 1, c' = c − 2. Then a' < b' < c' are distinct integers in {1, ..., 18}, and this map is a bijection. The count is C(18, 3) = 816. Check by complement: the total is C(20,3) = 1140. Selections with at least one adjacent pair number 19×18 − 18 = 324, since the 18 consecutive triples are counted twice in 19×18. So 1140 − 324 = 816.

Answer: (D)

LEVEL has 5 letters with L twice and E twice. The number of arrangements is 5! / (2! · 2!) = 120 / 4 = 30.

Answer: (A)

Split by the units digit, since 0 is special. If it is 0, the thousands digit has 9 choices, then 8, then 7 for the remaining places: 504. If it is one of 2, 4, 6, 8 (4 ways), the thousands digit cannot be 0 or the units digit, so it has 8 choices. The hundreds digit then has 8 choices and the tens digit 7: 4 × 8 × 8 × 7 = 1792. Total = 504 + 1792 = 2296.

Answer: (A)

Count the complement. Total triangles = C(12,3) = 220. Classify the triangles that contain at least one side of the dodecagon. (i) Exactly two polygon sides: the three vertices are consecutive, so there are 12 such triangles. (ii) Exactly one polygon side: choose the side in 12 ways. The third vertex must not be either neighbour of that side's endpoints, which removes 2 vertices besides the 2 endpoints. That leaves 12 − 4 = 8 choices, giving 12 × 8 = 96 triangles. Triangles with at least one polygon side = 12 + 96 = 108. Required number = 220 − 108 = 112. Check: choosing a side and any other vertex gives 12 × 10 = 120, which counts the 12 two-side triangles twice. 120 − 12 = 108, which agrees.

Answer: (D)

Take the letters one at a time. Each letter can be posted in any of the 3 boxes, and the choice for one letter does not affect the others because boxes can hold several letters. By the multiplication principle the total is 3 × 3 × 3 × 3 = 81. The answer 4³ = 64 would count giving each of 3 letters a choice among 4 boxes, which reverses the roles. Answers 12 and 24 treat the problem as a restricted selection rather than independent choices with repetition allowed.

Answer: (A)

Split by the number of women. Exactly 2: C(4,2)·C(5,2) = 6·10 = 60. Exactly 3: C(4,3)·C(5,1) = 4·5 = 20. Exactly 4: C(4,4) = 1. Total = 60 + 20 + 1 = 81. Check by complement: C(9,4) = 126, minus 0 women C(5,4) = 5, minus 1 woman 4·C(5,3) = 40, gives 81.

Answer: (A)

First count committees with at least 2 women. The total is C(11,5) = 462. Remove 0 women, C(6,5) = 6, and 1 woman, 5 × C(6,4) = 75. This leaves 462 − 81 = 381. Now remove the committees that contain both A and B and also satisfy the women condition. With A and B in, the other 3 members come from the remaining 4 men and 5 women, which is 9 people, with at least 2 women. C(9,3) = 84 in all, minus 0 women, C(4,3) = 4, minus 1 woman, 5 × C(4,2) = 30. This gives 50. The answer is 381 − 50 = 331.

Answer: (A)

C(8, 3) = (8 × 7 × 6)/6 = 56 and C(8, 4) = (8 × 7 × 6 × 5)/24 = 70. The sum is 126. This matches C(n, r) + C(n, r−1) = C(n+1, r) with n = 8, r = 4, giving C(9, 4) = 126.