Minimalist IAS
CSAT 2019 paper

UPSC CSE CSAT 2019 · Paper II (CSAT) · Question 14 · Counting, permutations & probability

चार समांतर रेखाओं के एक समुच्चय से, जो कि चार समांतर रेखाओं के एक अन्य समुच्चय से प्रतिच्छेदी है,…

UPSC Civil Services (Preliminary) Examination 2019, General Studies Paper II (CSAT), Question 14 — Counting, permutations & probability.

CSAT 2019 · Q14

Counting, permutations & probability Easy

चार समांतर रेखाओं के एक समुच्चय से, जो कि चार समांतर रेखाओं के एक अन्य समुच्चय से प्रतिच्छेदी है, बनाए जा सकने वाले समांतर चतुर्भुजों की संख्या है

Read this question in English

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of four parallel lines, is

(a) 18 · (b) 24 · (c) 32 · (d) 36

उत्तर और व्याख्या · explanation in English

Answer: (d) 36

Every choice of two lines from each family encloses exactly one parallelogram, so the count is 4C2 × 4C2 = 6 × 6 = 36.

  1. A parallelogram needs two lines from the first set and two from the second.
  2. Ways to choose 2 lines out of 4 = 4C2 = 6, for each set.
  3. Parallelograms = 6 × 6 = 36.

Remember · m parallel lines crossing n parallel lines form mC2 × nC2 parallelograms.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2019, Q14 — minimalistias.com/pyqs/csat/2019/q14

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

Timed · 35 minutes Take CSAT Counting, permutations & probability · Set 1 as a test This question is in it — with UPSC's marking and an instant score. Start the set

More on Counting, permutations & probability

All Counting, permutations & probability questions →