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?
Answer & explanation
Answer: (d) Four triplets
Since x ≤ 7, 2y ≤ 6 and so 3z < 6, which forces z = 1 and y = 2 or 3. That gives (5, 2, 1), (6, 2, 1), (7, 2, 1) and (7, 3, 1): four triplets.
- x ≤ 7 and x > 2y, so 2y ≤ 6, i.e. y ≤ 3.
- 2y > 3z with 2y ≤ 6 means 3z < 6, so z = 1; then 2y > 3 gives y = 2 or 3.
- y = 2: x > 4, so x = 5, 6 or 7, giving three triplets.
- y = 3: x > 6, so x = 7, giving one triplet.
- Total = 4 triplets: (5, 2, 1), (6, 2, 1), (7, 2, 1), (7, 3, 1).
Remember · In chained inequalities, start from the tightest bound (here x ≤ 7) and work down to the smallest variable.
Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·