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UPSC CSE CSAT 2022 · Paper II (CSAT) · Question 27 · Counting, permutations & probability

A, B और C तीन स्थान इस प्रकार हैं कि A से B के लिए तीन भिन्न रास्ते हैं, B से C के लिए चार भिन्न…

UPSC Civil Services (Preliminary) Examination 2022, General Studies Paper II (CSAT), Question 27 — Counting, permutations & probability.

CSAT 2022 · Q27

Counting, permutations & probability Easy

A, B और C तीन स्थान इस प्रकार हैं कि A से B के लिए तीन भिन्न रास्ते हैं, B से C के लिए चार भिन्न रास्ते हैं और A से C के लिए तीन भिन्न रास्ते हैं। इन रास्तों का प्रयोग कर कोई व्यक्ति कितने भिन्न मार्गों से A से C तक जा सकता है?

Read this question in English

A, B and C are three places such that there are three different roads from A to B, four different roads from B to C and three different roads from A to C. In how many different ways can one travel from A to C using these roads?

(a) 10 · (b) 13 · (c) 15 · (d) 36

उत्तर और व्याख्या · explanation in English

Answer: (c) 15

One can go directly from A to C (3 ways) or go through B (3 × 4 = 12 ways). These are alternatives, so they add: 3 + 12 = 15.

  1. Direct routes from A to C: 3.
  2. Routes through B: 3 roads from A to B × 4 roads from B to C = 12.
  3. Total = 3 + 12 = 15.

Remember · Legs taken one after another multiply; alternative routes add.

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, Q27 — minimalistias.com/pyqs/csat/2022/q27

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

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