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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 2017 · Q38

Easy

There are 4 horizontal and 4 vertical lines, parallel and equidistant to one another on a board. What is the maximum number of rectangles and squares that can be formed?

Answer & explanation

Answer: (c) 36

Every rectangle, squares included, is fixed by choosing 2 of the 4 horizontal lines and 2 of the 4 vertical lines. That gives 6 × 6 = 36.

  1. Ways to choose 2 horizontal lines from 4 = 4C2 = 6.
  2. Ways to choose 2 vertical lines from 4 = 6.
  3. Each pair of choices gives one rectangle: 6 × 6 = 36 (squares included).
  4. Check: the grid has 3 × 3 unit cells; squares = 9 + 4 + 1 = 14 and non-square rectangles = 22; 14 + 22 = 36.

Remember · Rectangles in a grid of m horizontal and n vertical lines = mC2 × nC2; the squares are already included.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q56

Easy

A bag contains 20 balls. 8 balls are green, 7 are white and 5 are red. What is the minimum number of balls that must be picked up from the bag blindfolded (without replacing any of it) to be assured of picking at least one ball of each colour?

Answer & explanation

Answer: (b) 16

In the worst case you pick all 8 green and all 7 white balls first — 15 balls and still no red. The next ball must be red, so 16 balls guarantee at least one of each colour.

  1. Worst case: the two largest colour groups come out first — 8 green + 7 white = 15 balls.
  2. Only red balls remain, so the 16th ball is red.
  3. 16 is enough in every case, and 15 is not (the all-green-and-white case), so the minimum is 16.

Remember · To be assured of one of each colour, take every ball except the smallest colour group, then add 1.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2017 · Q57

Easy

If 2 boys and 2 girls are to be arranged in a row so that the girls are not next to each other, how many possible arrangements are there?

Answer & explanation

Answer: (c) 12

There are 4! = 24 arrangements in all. Treating the two girls as one block gives 3! × 2 = 12 arrangements in which they sit together, so 24 − 12 = 12 keep the girls apart.

  1. All arrangements of 4 different children: 4! = 24.
  2. Girls together: treat them as one unit → 3 units → 3! = 6, and the girls can swap within the unit → 6 × 2 = 12.
  3. Girls not together: 24 − 12 = 12.
  4. Check (gap method): arrange the boys in 2! = 2 ways, leaving 3 gaps; place the two girls in 2 of the gaps in 3 × 2 = 6 ways; total 2 × 6 = 12.

Remember · 'Not together' = total − together; or seat the others first and put the separated people in the gaps.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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