Permutation, Combination & Probability MCQs for UPSC Prelims

We need three-digit numbers with strictly decreasing digits (hundreds > tens > units). The number 1000 has digits 1,0,0,0 and does not qualify, so the hundreds digit can only be 7, 8 or 9. For a fixed hundreds digit H, the tens and units digits must be two distinct digits both less than H, arranged in decreasing order; choosing any 2 distinct digits from {0,1,...,H-1} fixes their order, giving C(H, 2) numbers. For H = 7: C(7,2) = 21. For H = 8: C(8,2) = 28. For H = 9: C(9,2) = 36. Total = 21 + 28 + 36 = 85. Hence (c) 85.

Count distinct colourings of a cube with 2 colours where two colourings are the same if one can be rotated into the other. Group by number of black faces: 0 black (1 way), 6 black (1 way); 1 black (1 way, all single faces are equivalent by rotation), 5 black (1 way); 2 black (2 ways, the black faces are either adjacent or opposite), 4 black (2 ways, by symmetry with 2 black); 3 black (2 ways, the three black faces either share a common vertex or form a band/strip). Adding: 1 + 1 + 1 + 1 + 2 + 2 + 2 = 10. (This matches Burnside's count (1/24)(2^6 + 6*2^3 + 3*2^4 + 8*2^2 + 6*2^3) = 240/24 = 10.) So the answer is (b) 10.

To go from A (top-left) to B (bottom-right) of a 2 x 2 grid of cells, the person must make exactly 2 moves to the right (R) and 2 moves downward (D), in any order. The number of distinct sequences of these 4 moves is the number of ways to choose which 2 of the 4 steps are 'right', that is 4C2 = 4! / (2! x 2!) = 6. So there are 6 different routes.

99% of the 400 total is 396, so the four integer scores must add to 396, i.e. the shortfalls from 100 in the four papers must add to 4 with each shortfall between 0 and 100. The number of non-negative integer solutions of d1 + d2 + d3 + d4 = 4 is C(4 + 3, 3) = C(7, 3) = 35 (the cap of 100 never binds since each shortfall is at most 4). Enumerating score patterns confirms it: (100,100,100,96) in 4 ways, (100,100,99,97) in 12, (100,100,98,98) in 6, (100,99,99,98) in 12, (99,99,99,99) in 1, totalling 35. So (d) is correct; 16, 17 and 23 come from missing some of these distributions.

Any five-digit number formed from the digits 1, 2, 3, 4, 5 uses each digit exactly once, so its digit sum is always 1 + 2 + 3 + 4 + 5 = 15. Since 15 is divisible by 3, every such arrangement is divisible by 3 (and is greater than 3), so none can be prime. Therefore the count of primes is Zero.

Choose the Principal first. Only 2 of the 6 candidates are eligible for Principal, so there are 2 ways. Once the Principal is fixed, that person is no longer available, leaving 5 candidates for the two identical Vice-Principal posts. The two Vice-Principals are an unordered pair, so the count is 5C2 = 10. Total combinations = 2 x 10 = 20. Since 20 does not appear among options (a) 4, (b) 12, or (c) 18, the correct choice is None of the above. The official UPSC 2015 key marks this answer as (d).

A number divisible by 5 must end in 0 or 5; since every digit must be odd, the units digit is 5. The hundreds and tens digits must come from the remaining odd digits {1, 3, 7, 9} without repetition: 4 choices for the hundreds place and 3 for the tens place, giving 4 x 3 = 12 numbers, option (b). 8 undercounts the choices, while 16 and 24 ignore the no-repetition or odd-digit constraints.

Number the six persons 1 to 6 in the row. We need 3-element subsets with no two consecutive numbers. The count of ways to pick k non-consecutive items from n in a row is C(n-k+1, k) = C(4, 3) = 4. Listing them confirms it: {1,3,5}, {1,3,6}, {1,4,6}, {2,4,6}. So the answer is 4, option (b). Option (a) 3 misses one valid set, while option (c) 5 and option (d) 6 include sets such as {1,2,4} or {2,4,5} that contain adjacent persons.