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UPSC CSE CSAT 2022 · Paper II (CSAT) · Question 24 · Counting, permutations & probability

वर्ण A, B, C, D और E इस तरह व्यवस्थित किए गए हैं कि A और E के बीच यथातथ्य दो वर्ण हैं। इस तरह…

UPSC Civil Services (Preliminary) Examination 2022, General Studies Paper II (CSAT), Question 24 — Counting, permutations & probability.

CSAT 2022 · Q24

Counting, permutations & probability Easy

वर्ण A, B, C, D और E इस तरह व्यवस्थित किए गए हैं कि A और E के बीच यथातथ्य दो वर्ण हैं। इस तरह कितनी व्यवस्थाएँ संभव हैं?

Read this question in English

The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?

(a) 12 · (b) 18 · (c) 24 · (d) 36

उत्तर और व्याख्या · explanation in English

Answer: (c) 24

In a row of five, A and E can stand only at places 1 and 4 or 2 and 5, in either order: 4 ways. The other three letters then fill the remaining places in 3! = 6 ways, giving 24.

  1. Number the places 1 to 5. Exactly two letters between A and E means A and E occupy places (1, 4) or (2, 5): 2 choices of places.
  2. A and E can swap within the chosen places: 2 × 2 = 4 ways.
  3. B, C and D fill the other three places in 3! = 6 ways.
  4. Total = 4 × 6 = 24.

Remember · Place the constrained pair first (positions × internal order), then arrange the remaining items freely.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2022, Q24 — minimalistias.com/pyqs/csat/2022/q24

In the UPSC syllabus: Counting, permutations & probability › Counting figures & cases See it and tick it covered →

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