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CSAT 2021 paper

UPSC CSE CSAT 2021 · Paper II (CSAT) · Question 69 · Counting, permutations & probability

There are 6 persons arranged in a row. Another person has to shake hands with 3 of them so that he…

UPSC Civil Services (Preliminary) Examination 2021, General Studies Paper II (CSAT), Question 69 — Counting, permutations & probability.

CSAT 2021 · Q69

Counting, permutations & probability 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). ·

Cite this question: UPSC CSE CSAT 2021, Q69 — minimalistias.com/pyqs/csat/2021/q69

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