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UPSC CSE CSAT 2020 · Paper II (CSAT) · Question 67 · Counting, permutations & probability

How many different 5-letter words (with or without meaning) can be constructed using all the…

UPSC Civil Services (Preliminary) Examination 2020, General Studies Paper II (CSAT), Question 67 — Counting, permutations & probability.

CSAT 2020 · Q67

Counting, permutations & probability Easy

How many different 5-letter words (with or without meaning) can be constructed using all the letters of the word ‘DELHI’ so that each word has to start with D and end with I?

Answer & explanation

Answer: (d) 6

With D fixed first and I fixed last, only E, L and H are left to arrange in the three middle places. Three distinct letters can be arranged in 3! = 6 ways.

  1. Positions 1 and 5 are fixed: D _ _ _ I.
  2. E, L and H fill the three middle places in 3! = 3 × 2 × 1 = 6 ways.
  3. Check: DELHI, DEHLI, DLEHI, DLHEI, DHELI, DHLEI — six words.

Remember · Fix the letters whose places are given, then arrange only the remaining ones: n distinct letters give n! orders.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2020, Q67 — minimalistias.com/pyqs/csat/2020/q67

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