Minimalist IAS
Sign up
English हिंदी
CSAT

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)
Timed · 30 minutes a set Practise Counting, permutations & probability as timed tests 2 sets of about 20 questions with UPSC's CSAT marking. Start practising

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);” Sub-topics in the detailed syllabus →

CSAT 2017 · Q38

Easy

एक पट्ट पर 4 क्षैतिज और 4 ऊर्ध्वाधर रेखाएँ हैं, जो परस्पर समान्तर और समदूरस्थ हैं। इनसे अधिकतम कितने आयत और वर्ग बनाए जा सकते हैं?

Read this question in English

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?

(a) 16 · (b) 24 · (c) 36 · (d) 42

उत्तर और व्याख्या · explanation in English

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

एक थैले में 20 गेंदें हैं। 8 गेंदें हरी हैं, 7 सफ़ेद हैं और 5 लाल हैं। आँख बंद कर, थैले में से न्यूनतम कितनी गेंदें निकालना आवश्यक है (किसी को भी बिना बदले) जिससे सुनिश्चित हो कि प्रत्येक रंग की कम-से-कम एक गेंद निकली हो?

Read this question in English

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?

(a) 17 · (b) 16 · (c) 13 · (d) 11

उत्तर और व्याख्या · explanation in English

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

यदि 2 लड़कों और 2 लड़कियों को एक पंक्ति में इस व्यवस्था में खड़ा करना हो कि लड़कियाँ एक-दूसरे के अगल-बगल खड़ी न हों, तो कितनी संभव व्यवस्थाएँ हो सकती हैं?

Read this question in English

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?

(a) 3 · (b) 6 · (c) 12 · (d) 24

उत्तर और व्याख्या · explanation in English

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 ·

The Syllabus X-Ray of this topic

1 of its 3 sub-topics is High

0 Core · 1 High · 2 Steady · 0 Rare

  • HighCounting figures & cases · 37 questions

The X-Ray ranks all 928 sub-topics of the UPSC syllabus by what UPSC has asked since 2016. The 6 CSAT Core sub-topics are behind 50% of CSAT questions.

See the whole X-Ray →

Other CSAT skills