Counting, permutations & probability

Answer: (d) 36

Every choice of two lines from each family encloses exactly one parallelogram, so the count is 4C2 × 4C2 = 6 × 6 = 36.

1. A parallelogram needs two lines from the first set and two from the second.
2. Ways to choose 2 lines out of 4 = 4C2 = 6, for each set.
3. Parallelograms = 6 × 6 = 36.

Remember · m parallel lines crossing n parallel lines form mC2 × nC2 parallelograms.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 18

Count by the number of ₹50 notes. With none, ₹107 can include 0 to 10 ten-rupee notes (11 ways); with one, ₹57 can include 0 to 5 (6 ways); with two, ₹7 uses only ₹1 coins (1 way). Total 18.

1. No ₹50 note: ₹107 = 10a + b with a = 0, 1, …, 10 → 11 ways.
2. One ₹50 note: ₹57 = 10a + b with a = 0, 1, …, 5 → 6 ways.
3. Two ₹50 notes: ₹7 = 10a + b needs a = 0 → 1 way.
4. In each case ₹1 coins make up the balance, so each choice is one way.
5. Total = 11 + 6 + 1 = 18.

Remember · For currency combinations, fix the largest denomination, count choices for the next; the smallest always fills the rest.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 10

Treating paintings that can be turned into one another as the same, count by the number of black faces: 0, 1, 5 or 6 black faces give one pattern each, while 2, 3 or 4 give two each. Total 1 + 1 + 2 + 2 + 2 + 1 + 1 = 10.

1. Count by the number of black faces; rotations of the cube give the same painting.
2. 0 black: 1 way. 1 black: 1 way (any face can be turned to the top).
3. 2 black: adjacent or opposite → 2 ways.
4. 3 black: all three meeting at a corner, or in a band (two opposite faces plus one between) → 2 ways.
5. 4 black = 2 white → 2 ways; 5 black → 1 way; 6 black → 1 way.
6. Total = 1 + 1 + 2 + 2 + 2 + 1 + 1 = 10.

Remember · For two-colour cube patterns, count by faces of one colour; k and 6 − k faces give the same count.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 10

Each of x, y and z is at least 1, so we split 6 units into three positive parts: 5C2 = 10 ordered triplets.

1. Natural numbers start from 1, so x, y, z ≥ 1.
2. Line up 6 units and choose 2 of the 5 gaps between them to cut into three positive parts: 5C2 = 10.
3. Check by listing: (4, 1, 1) in 3 orders, (3, 2, 1) in 6 orders, (2, 2, 2) in 1 order → 3 + 6 + 1 = 10.

Remember · Positive solutions of x₁ + … + xₖ = n number (n − 1)C(k − 1); order matters in triplets.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error