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UPSC CSE CSAT 2019 · Paper II (CSAT) · Question 73 · Counting, permutations & probability

How many triplets (x, y, z) satisfy the equation x + y + z = 6, where x, y and z are natural…

UPSC Civil Services (Preliminary) Examination 2019, General Studies Paper II (CSAT), Question 73 — Counting, permutations & probability.

CSAT 2019 · Q73

Counting, permutations & probability Easy

How many triplets (x, y, z) satisfy the equation x + y + z = 6, where x, y and z are natural numbers?

Answer & explanation

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). ·

Cite this question: UPSC CSE CSAT 2019, Q73 — minimalistias.com/pyqs/csat/2019/q73

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