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?
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A hard CSAT / General Aptitude multiple-choice question from UPSC Prelims 2024. Attempt it above, then check the correct answer and detailed explanation below.
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Correct answer: D. Four triplets
From {1,…,7} with x > 2y > 3z: z must be 1 (else 3z ≥ 6 forces 2y > 6, y ≥ 4, x > 8 impossible). With z=1: y=2 gives x ∈ {5,6,7} and y=3 gives x=7 — four triplets: (5,2,1),(6,2,1),(7,2,1),(7,3,1). So option (d).
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