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UPSC CSE CSAT 2018 · Paper II (CSAT) · Question 62 · Counting, permutations & probability

For a sports meet, a winners’ stand comprising three wooden blocks is in the following form:

UPSC Civil Services (Preliminary) Examination 2018, General Studies Paper II (CSAT), Question 62 — Counting, permutations & probability.

CSAT 2018 · Q62

Counting, permutations & probability Easy

For a sports meet, a winners’ stand comprising three wooden blocks is in the following form:

A winners' stand of three rectangular blocks side by side: the tallest on the left, a middle-height block next to it and the shortest on the right.
From UPSC's question paper.

There are six different colours available to choose from and each of the three wooden blocks is to be painted such that no two of them has the same colour. In how many different ways can the winners’ stand be painted?

Answer & explanation

Answer: (a) 120

The three blocks are different (first, second and third place), so the order of colours matters. With six colours and no repeats there are 6 × 5 × 4 = 120 ways.

  1. The blocks differ in height, so each assignment of colours to blocks is a different painting.
  2. Tallest block: 6 choices; next block: 5; last block: 4.
  3. Ways = 6 × 5 × 4 = 120 (that is, ⁶P₃).

Remember · When the positions are distinct, use permutations: n × (n − 1) × … for as many positions as are filled.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 1 Oct 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2018, Q62 — minimalistias.com/pyqs/csat/2018/q62

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