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UPSC CSE CSAT 2023 · Paper II (CSAT) · Question 30 · Counting, permutations & probability

ABCD कोई वर्ग है । AB और CD प्रत्येक पर एक बिंदु; और BC और DA प्रत्येक पर दो भिन्न बिंदु चुने जाते…

UPSC Civil Services (Preliminary) Examination 2023, General Studies Paper II (CSAT), Question 30 — Counting, permutations & probability.

CSAT 2023 · Q30

Counting, permutations & probability Easy

ABCD कोई वर्ग है । AB और CD प्रत्येक पर एक बिंदु; और BC और DA प्रत्येक पर दो भिन्न बिंदु चुने जाते हैं । इन छह बिन्दुओं में से किन्हीं तीन बिन्दुओं को शीर्ष ले कर कितने भिन्न त्रिभुज खींचे जा सकते हैं ?

Read this question in English

ABCD is a square. One point on each of AB and CD; and two distinct points on each of BC and DA are chosen. How many distinct triangles can be drawn using any three points as vertices out of these six points?

(a) 16 · (b) 18 · (c) 20 · (d) 24

उत्तर और व्याख्या · explanation in English

Answer: (c) 20

Three points fail to make a triangle only if they lie on one straight line. No side of the square carries more than two of the six points, so no three are in a line and every choice of three points gives a triangle: C(6, 3) = 20.

  1. Points: 1 on AB, 1 on CD, 2 on BC and 2 on DA — six in all.
  2. A set of three points forms no triangle only when all three lie on one side; no side has more than two points.
  3. Number of triangles = C(6, 3) = (6 × 5 × 4)/(3 × 2 × 1) = 20.

Remember · Triangles from points = C(n, 3) minus collinear triples; check each line for three or more points.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2023, Q30 — minimalistias.com/pyqs/csat/2023/q30

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