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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 2016 · Q40

Easy

In a question paper there are five questions to be attempted and answer to each question has two choices – True (T) or False (F). It is given that no two candidates have given the answers to the five questions in an identical sequence. For this to happen the maximum number of candidates is:

Answer & explanation

Answer: (d) 32

Each question can be answered in 2 ways, so there are 2⁵ = 32 different answer sequences. If no two candidates repeat a sequence, at most 32 candidates are possible.

  1. Each of the 5 questions has 2 possible answers.
  2. Distinct sequences = 2 × 2 × 2 × 2 × 2 = 2⁵ = 32.
  3. No two candidates share a sequence, so the maximum is 32.

Remember · Independent choices multiply: k options at each of n places give kⁿ sequences.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2016 · Q63

Easy

How many numbers are there between 100 and 300 which either begin with or end with 2?

Answer & explanation

Answer: (a) 110

All numbers from 200 to 299 begin with 2 — 100 numbers — and 20 numbers from 102 to 292 end in 2. Ten of those (202 to 292) are counted twice, so the total is 100 + 20 − 10 = 110.

  1. Beginning with 2: 200 to 299 — 100 numbers.
  2. Ending in 2: 102, 112, …, 292 — 20 numbers.
  3. Both: 202, 212, …, 292 — 10 numbers.
  4. Total = 100 + 20 − 10 = 110.

Remember · 'Either … or' counting: add both groups, then subtract the overlap once.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2016 · Q72

Medium

Four-digit numbers are to be formed using the digits 1, 2, 3 and 4; and none of these four digits are repeated in any manner. Further,

  1. 1.2 and 3 are not to immediately follow each other
  2. 2.1 is not to be immediately followed by 3
  3. 3.4 is not to appear at the last place
  4. 4.1 is not to appear at the first place

How many different numbers can be formed?

Answer & explanation

Answer: (a) 6

Listing the arrangements by first digit (1 cannot lead) and striking out any with 2 and 3 side by side, with 1 directly before 3, or with 4 at the end leaves 2143, 2431, 3142, 3412, 3421 and 4312 — six numbers.

  1. 1 cannot come first, so the number starts with 2, 3 or 4.
  2. Starting with 2: 2143 and 2431 work (2134 and 2413 contain 13; 2314 and 2341 contain 23).
  3. Starting with 3: 3142, 3412 and 3421 work (3124 ends in 4; 3214 and 3241 contain 32).
  4. Starting with 4: only 4312 works (4123 and 4231 contain 23; 4132 and 4213 contain 13; 4321 contains 32).
  5. Total = 2 + 3 + 1 = 6.

Remember · With only 24 arrangements, list them systematically by first digit and strike out rule-breakers rather than trusting a formula with overlapping conditions.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2016 · Q79

Easy

A round archery target of diameter 1 m is marked with four scoring regions from the centre outwards as red, blue, yellow and white. The radius of the red band is 0.20 m. The width of all the remaining bands is equal. If archers throw arrows towards the target, what is the probability that the arrows fall in the red region of the archery target?

Answer & explanation

Answer: (c) 0.16

Taking every point of the target as equally likely, the probability is the red area over the whole area: (0.20 ÷ 0.50)² = 0.16. The widths of the other bands do not matter.

  1. Target radius = 1 m ÷ 2 = 0.50 m; red radius = 0.20 m.
  2. Probability = red area ÷ target area = (0.20/0.50)² = 0.4² = 0.16.

Remember · For circles, the ratio of areas is the square of the ratio of radii.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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