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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)
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 2026 · Q9

Easy Provisional key

QUEUE शब्द के वर्णों का स्थान अदल-बदल कर कितने शब्द बनाए जा सकते हैं, यदि Q के बाद सदैव U आता है? इस प्रकार बने शब्द का कोई अर्थ होना आवश्यक नहीं है।

Read this question in English

How many words can one form by shuffling the letters of the word QUEUE, if Q is always followed by U? The words thus formed need not necessarily have any meaning.

(a) 6 · (b) 8 · (c) 10 · (d) 12

उत्तर और व्याख्या · explanation in English

Answer: (d) 12

Glue Q and U together as one block 'QU'. Then arrange four units — QU, U, E, E — where the two E's are identical: 4!/2! = 12.

  1. QUEUE has Q, U, U, E, E.
  2. Tie Q to a U as a block: the units are QU, U, E, E.
  3. Arrangements = 4! ÷ 2! (two identical E's) = 24 ÷ 2 = 12.
  4. Check: Q occurs once, so each word has exactly one block position — no word is counted twice.

Remember · 'X always followed by Y' → treat XY as one block, then divide by factorials of repeated letters.

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

CSAT 2026 · Q15

Medium Provisional key

एक मेज़ का ऊपरी पृष्ठ (top of a table) आयताकार है और उसके आयाम 6′ × 10′ हैं। मेज़ के ऊपरी पृष्ठ के दो आयताकार भागों को नीले रंग से रंगा गया है; इन दोनों भागों के आयाम 2·5′ × 8′ हैं और उनमें से प्रत्येक की सटीक दो भुजाएँ, मेज़ के ऊपरी पृष्ठ के दो किनारों के उभयनिष्ठ हैं। यदि मेज़ को ज़मीन पर स्थिर रखा जाता है और मेज़ के ऊपरी पृष्ठ के शेष भाग को सफ़ेद रंग से रंगा जाता है, तो ऊपर से देखने पर कितने भिन्न प्रतिरूप (पैटर्न) संभव हैं?

Read this question in English

The top of a table is rectangular and its dimensions are 6′ × 10′. Two rectangular portions of the table top are painted in blue colour; both these portions have dimensions 2.5′ × 8′ and each of them has exactly two sides common with two edges of the table top. If the table is fixed to the ground and the remaining portion of the table top is painted in white, how many different patterns are possible when observed from above?

(a) 2 · (b) 4 · (c) 6 · (d) 8

उत्तर और व्याख्या · explanation in English

Answer: (b) 4

Each 2·5′ × 8′ strip must lie with its 8′ side along a 10′ edge and sit in a corner. Two strips cannot share the same 10′ edge without overlapping, so one strip goes to each long edge, at its left or right end: 2 × 2 = 4 patterns.

  1. The 8′ side cannot lie along the 6′ edge (8 > 6), so it lies along a 10′ edge and the 2·5′ side along a 6′ edge.
  2. Two sides common with two table edges means the strip sits in a corner (it cannot touch two opposite edges, since 2·5 < 6 and 8 < 10).
  3. Two strips along the same 10′ edge would overlap (8 + 8 > 10), so one strip lies along each 10′ edge; they do not meet since 2·5 + 2·5 = 5 < 6.
  4. Each strip can be at the left or the right end of its edge: 2 × 2 = 4.
  5. The table is fixed, so these four layouts (both left, both right, and the two diagonal ones) all look different from above.

Remember · Fix the orientation first (which side fits which edge), then count positions; a fixed table means mirror images count separately.

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

CSAT 2026 · Q23

Medium Provisional key

चार प्रकार के बाट (weights) हैं, नामतः 1 kg, 2 kg, 5 kg और 10 kg. यदि मापते समय, कम-से-कम आठ, किन्तु ग्यारह से अनधिक 1 kg के बाटों का उपयोग किया जाना है, तो 20 kg के वज़न को अधिकतम कितने भिन्न प्रकार से मापा जा सकता है?

Read this question in English

There are four types of weights, namely 1 kg, 2 kg, 5 kg and 10 kg. What is the maximum number of different ways one can measure 20 kg, if at least eight but not more than eleven weights of 1 kg are to be used while measuring?

(a) 7 · (b) 8 · (c) 9 · (d) 10

उत्तर और व्याख्या · explanation in English

Answer: (b) 8

Fix the number of 1 kg weights at 8, 9, 10 or 11 and make up the rest with 2, 5 and 10 kg weights. The counts are 3, 1, 3 and 1, giving 8 ways.

  1. Eight 1 kg weights: 12 kg left → 10 + 2; 5 + 5 + 2; six 2s → 3 ways.
  2. Nine 1 kg weights: 11 kg left from 2, 5 and 10 kg weights. 11 is odd, so exactly one 5 is needed, plus three 2s → 1 way.
  3. Ten 1 kg weights: 10 kg left → 10; 5 + 5; five 2s → 3 ways.
  4. Eleven 1 kg weights: 9 kg left, odd again → 5 + 2 + 2 → 1 way.
  5. Total = 3 + 1 + 3 + 1 = 8.

Remember · Counting combinations of coins or weights: fix the most restricted item, then list the rest systematically from the largest.

Question and answer: UPSC's provisional GS Paper II (2026, 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.

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