Permutations and Combinations: JEE Advanced PYQ, 2007–2026

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: (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: (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)

The number is greater than 3000, so the thousands digit is 3, 4 or 5. The last digit must be even, from {0, 2, 4}, and different from the thousands digit. Thousands digit 3: units digit has 3 choices, and the middle two places have 4×3 = 12 ways, so 36. Thousands digit 4: units digit is 0 or 2, 2 choices, so 2×12 = 24. Thousands digit 5: units digit has 3 choices, so 36. Total = 36 + 24 + 36 = 96.

Answer: (A)

A product of digits is divisible by 10 in exactly two disjoint situations. Case 1: some digit is 0, so the product is 0. The first digit has 9 choices, so there are 9000 four-digit numbers. Those with no 0 number 9^4 = 6561, so 9000 − 6561 = 2439 contain a 0. Case 2: no digit is 0, so every digit is in {1,...,9}. The product then needs a factor 5 and a factor 2, so the number must contain a 5 and at least one even digit from {2,4,6,8}. Count the strings in {1..9}^4 as total − (no 5) − (no even digit) + (neither). This gives 9^4 − 8^4 − 5^4 + 4^4 = 6561 − 4096 − 625 + 256 = 2096. The two cases are disjoint, so the total is 2439 + 2096 = 4535.

Answer: (A)

When identical letters must be kept apart, place the other letters first and then slot the restricted letters into the gaps. The letters U, C, C, E give 4!/2! = 12 arrangements. They leave 5 gaps (4 internal and 2 ends, counted as 5 positions in all). The three identical S's go into 3 different gaps, so choose C(5,3) = 10. The total is 12 × 10 = 120.

Answer: (A)

At least 2 women: C(4,2)C(6,3) + C(4,3)C(6,2) + C(4,4)C(6,1) = 120 + 60 + 6 = 186. Now remove committees containing both M and W. Fix M and W and choose 3 more from the remaining 5 men and 3 women. W is already one woman, so at least one more woman is needed. This gives C(8,3) − C(5,3) = 56 − 10 = 46. Answer = 186 − 46 = 140.