Minimalist IAS
CSAT

CSAT · 43 questions

Counting, permutations & probability

Every UPSC CSAT question on this topic, 2016–2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Counting, permutations & probability questions per year: 2016: 4, 2017: 3, 2018: 4, 2019: 4, 2020: 2, 2021: 3, 2022: 8, 2023: 11, 2024: 1, 2025: 0, 2026: 3 Asked in 10 of 11 years · most in 2023 (11)

UPSC syllabus: “Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc. — Class X level);” See the full syllabus →

CSAT 2024 · Q49

Medium

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.

  1. x ≤ 7 and x > 2y, so 2y ≤ 6, i.e. y ≤ 3.
  2. 2y > 3z with 2y ≤ 6 means 3z < 6, so z = 1; then 2y > 3 gives y = 2 or 3.
  3. y = 2: x > 4, so x = 5, 6 or 7, giving three triplets.
  4. y = 3: x > 6, so x = 7, giving one triplet.
  5. 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). Permalink ·

Other CSAT skills