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UPSC CSE CSAT 2021 · Paper II (CSAT) · Question 66 · Counting, permutations & probability

Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed?

UPSC Civil Services (Preliminary) Examination 2021, General Studies Paper II (CSAT), Question 66 — Counting, permutations & probability.

CSAT 2021 · Q66

Counting, permutations & probability Easy

Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed?

Answer & explanation

Answer: (b) 6

A five-digit number from these digits exceeds 30000 only if it starts with 3. The remaining digits 2, 2, 3, 3 can be arranged in 4!/(2! × 2!) = 6 distinct ways.

  1. A number made from 2, 2, 3, 3, 3 is greater than 30000 only if its first digit is 3.
  2. The remaining digits 2, 2, 3, 3 can be arranged in 4! ÷ (2! × 2!) = 24 ÷ 4 = 6 ways.
  3. So 6 distinct numbers are greater than 30000.
  4. Check: 32233, 32323, 32332, 33223, 33232, 33322.

Remember · Arrangements with repeated items = n! ÷ (product of the factorials of the repeat counts).

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2021, Q66 — minimalistias.com/pyqs/csat/2021/q66

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