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UPSC CSE CSAT 2017 · Paper II (CSAT) · Question 57 · Counting, permutations & probability

यदि 2 लड़कों और 2 लड़कियों को एक पंक्ति में इस व्यवस्था में खड़ा करना हो कि लड़कियाँ एक-दूसरे के…

UPSC Civil Services (Preliminary) Examination 2017, General Studies Paper II (CSAT), Question 57 — Counting, permutations & probability.

CSAT 2017 · Q57

Counting, permutations & probability 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). ·

Cite this question: UPSC CSE CSAT 2017, Q57 — minimalistias.com/pyqs/csat/2017/q57

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