UPSC Prelims 2020·CSAT·Quantitative Aptitude·Combinatorics and Probability

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?

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Last updated 23 May 2026, 3:31 pm IST
  1. A24
  2. B18
  3. C12
  4. D6Correct

Explanation

The word 'DELHI' has 5 distinct letters: D, E, L, H, I. We need to construct 5-letter words using all these letters. The constraints are: 1. The word must start with D. 2. The word must end with I. Let's represent the 5-letter word with 5 blanks: _ _ _ _ _ According to the constraints: The first position is fixed as D. D _ _ _ _ The last position is fixed as I. D _ _ _ I Now, we have 3 remaining letters (E, L, H) to fill the 3 middle positions. The number of ways to arrange these 3 distinct letters in the 3 remaining positions is 3 factorial (3!). 3! = 3 x 2 x 1 = 6. Therefore, there are 6 different 5-letter words that can be constructed under the given conditions. Analyzing the options: A) 24: This would be the case if we were arranging 4 letters (4!) or if we were arranging all 5 letters without any constraints (5! = 120) and then dividing by something, which is not applicable here. B) 18: Incorrect. C) 12: Incorrect. D) 6: This matches our calculated value (3!). The final answer is D.
Quantitative Aptitude: How many different 5-letter words (with or without meaning) can be constructed using all the letters of the word 'DELHI'

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