Minimalist IAS
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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 2021 · Q64

Medium

On a chess board, in how many different ways can 6 consecutive squares be chosen on the diagonals along a straight path?

Answer & explanation

Answer: (b) 6

The diagonals of a chess board are its two main diagonals of 8 squares each. A run of 6 consecutive squares fits in 8 − 6 + 1 = 3 positions on each, giving 6 ways.

  1. 'The diagonals' of a chess board are the two main diagonals, each 8 squares long.
  2. On a line of 8 squares, 6 consecutive squares can start at square 1, 2 or 3 — 8 − 6 + 1 = 3 ways.
  3. Two diagonals: 3 × 2 = 6 ways.
  4. Note: counting every diagonal line of 6 or more squares would give 18, which is not an option — confirming the main-diagonal reading.

Remember · Runs of k consecutive cells in a line of n cells = n − k + 1.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2021 · Q66

Easy

Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed?

Answer & explanation

Answer: (b) 6

A five-digit number from these digits exceeds 30000 only if it starts with 3. The remaining digits 2, 2, 3, 3 can be arranged in 4!/(2! × 2!) = 6 distinct ways.

  1. A number made from 2, 2, 3, 3, 3 is greater than 30000 only if its first digit is 3.
  2. The remaining digits 2, 2, 3, 3 can be arranged in 4! ÷ (2! × 2!) = 24 ÷ 4 = 6 ways.
  3. So 6 distinct numbers are greater than 30000.
  4. Check: 32233, 32323, 32332, 33223, 33232, 33322.

Remember · Arrangements with repeated items = n! ÷ (product of the factorials of the repeat counts).

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2021 · Q69

Easy

There are 6 persons arranged in a row. Another person has to shake hands with 3 of them so that he should not shake hands with two consecutive persons. In how many distinct possible combinations can the handshakes take place?

Answer & explanation

Answer: (b) 4

Choosing 3 of 6 people in a row with no two next to each other leaves only {1, 3, 5}, {1, 3, 6}, {1, 4, 6} and {2, 4, 6}. That is 4 combinations, matching C(4, 3).

  1. Number the persons 1 to 6 and choose 3 with no two adjacent.
  2. Possible sets: {1, 3, 5}, {1, 3, 6}, {1, 4, 6}, {2, 4, 6}.
  3. So there are 4 combinations.
  4. Check: C(n − k + 1, k) = C(6 − 3 + 1, 3) = C(4, 3) = 4.

Remember · Ways to pick k non-adjacent items from n in a row = C(n − k + 1, k).

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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