Counting & Combinatorial Reasoning MCQs for UPSC Prelims

Correct answer: D - 45

Option (d) is correct. A wearable pair means a left shoe and a right shoe of the same colour. The box has 54 shoes: 20 red (10 left, 10 right), 18 white and 16 black. In the worst case Raj first draws all 18 white and all 16 black shoes (34 attempts), and then 10 red shoes that are all for the same foot, say all 10 left ones (44 attempts), still without a wearable red pair. The 45th attempt must produce a right red shoe and complete the pair, so the maximum number of attempts is 45. Option (a) 27 is just one shoe from each pair and guarantees nothing. Option (b) 36 would be the answer only if any two red shoes counted as a pair (34 non-red plus 2 red); some coaching keys initially reported 36 on that reading, but a pair of shoes to wear requires a left and a right, and the official UPSC answer key confirms 45. Option (c) 44 is the worst case that still lacks a pair; one more attempt is needed to guarantee it.

Correct answer: A - 3

The figure has eight outer petal regions arranged in a ring around one central region. Going around the ring, alternate petals do not touch each other, so the petals can be coloured with just two colours used alternately (colour 1, colour 2, colour 1, colour 2, ...); since there is an even number of petals this alternation is consistent. The central region touches every petal, so it must take a third colour distinct from both. Therefore the minimum number of colours needed is 3.

Correct answer: B - 15

The k-th pass moves the ring k places clockwise, so after k passes the ring sits at position 1 + (1 + 2 + ... + k) = 1 + k(k+1)/2, taken modulo 40. The ring returns to child-1 when k(k+1)/2 is a multiple of 40, i.e. k(k+1) is a multiple of 80. Testing: k = 14 gives 210 (no), k = 15 gives 240 = 3 x 80 (yes). So after 15 changes the ring is back with child-1 - option (b), as per the official key. Option (c) 16 is the trap of counting the 16 holders (including child-1 at the start) instead of the 15 passes; options (a) and (d) fail the divisibility test.

Correct answer: C - 24

Exactly two letters between A and E means A and E occupy position pairs (1, 4) or (2, 5), and A and E can swap within each pair, giving 2 x 2 = 4 placements. The remaining letters B, C and D fill the three free positions in 3! = 6 ways. Total arrangements = 4 x 6 = 24, option (c). 12 misses the A-E swap, 18 has no valid construction, and 36 would require three position pairs when only two exist in a row of five.

Correct answer: D - 4

Out of F the only usable passage is the two-way F-D (B to F enters F but cannot be used outward). So every path starts F -> D. From D the options are E (two-way E-D) or H (D to H). Branch D -> E: from E we can go to A, C, or G. (1) E -> A reaches A: F-D-E-A. (2) E -> C, then C -> A: F-D-E-C-A. (3) E -> G, then G -> C, then C -> A: F-D-E-G-C-A (G-B is a dead end since B only leads back to F). Branch D -> H: H -> G, then G -> C, then C -> A: (4) F-D-H-G-C-A (the G-B option dead-ends at B). That gives 4 distinct simple paths. Answer: (d).

Correct answer: A - 1 only

The cuboid yields 7 x 5 x 3 = 105 unit cubes. Statement 1: unpainted cubes form the inner block (7-2) x (5-2) x (3-2) = 5 x 3 x 1 = 15, so it is correct. Statement 2: green sits on the 5 x 3 faces and blue on the 7 x 3 faces, so a cube showing exactly green and blue must lie on an edge where those faces meet - the four vertical edges of length 3. Each such edge has 3 cubes, but its two end cubes are corners that also touch a red 7 x 5 face (three painted faces), leaving 3 - 2 = 1 qualifying cube per edge, i.e. 4 cubes in total, not 6. Statement 2 is wrong. Hence option (a); (b) and (c) accept the miscount of 6, and (d) wrongly rejects the straightforward interior count of 15.

Correct answer: A - 27

Option (a) is correct. 125 identical cubes form a 5 x 5 x 5 block. A small cube is surrounded on every side exactly when it does not touch any outer face of the block, i.e. it lies in the inner core obtained by stripping one layer from each side. That core measures (5 - 2) x (5 - 2) x (5 - 2) = 3 x 3 x 3 = 27 cubes. Option (b) 25 is a face count, not a volume; options (c) 21 and (d) 18 correspond to no consistent counting of the hidden core.

Correct answer: B - 20

The solid is drawn as a regular icosahedron-style figure whose surface is made up entirely of triangular faces. Counting systematically by layers: the top cap contributes 5 triangles, the bottom cap contributes another 5 triangles, and the middle band between the two caps shows 10 triangles formed by the visible and dashed edges. Adding these gives 5 + 5 + 10 = 20 triangles. Hence the answer is 20.