Three numbers x, y, z are selected from the set of the first seven natural numbers such that x > 2y > 3z. How many such distinct triplets (x, y, z) are possible?
- (a) One triplet
- (b) Two triplets
- (c) Three triplets
- (d) Four triplets ✓ UPSC's answer
Why the answer is (d)
• The numbers come from {1, 2, …, 7} and must satisfy x > 2y > 3z.
• z = 1: 2y > 3 means y ≥ 2.
• y = 2: x > 4, so x = 5, 6, 7 → (5, 2, 1), (6, 2, 1), (7, 2, 1).
• y = 3: x > 6, so x = 7 → (7, 3, 1).
• y ≥ 4 needs x > 8 — impossible.
• z = 2: 2y > 6 means y ≥ 4, which again needs x > 8 — impossible. Larger z is impossible too.
• So there are four triplets — option (d).
Why the other options are wrong
- (a) One triplet
- There are four valid triplets, not one.
- (b) Two triplets
- There are four valid triplets, not two.
- (c) Three triplets
- There are four valid triplets, not three.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2024, held on 16 June 2024. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.