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?
- (a) 24
- (b) 18
- (c) 12
- (d) 6 ✓ UPSC's answer
Why the answer is (d)
• The word 'DELHI' consists of 5 distinct letters: D, E, L, H, I.
• The condition requires the first letter to be D and the last letter to be I, fixing the positions of these two letters.
• This leaves the middle three positions to be filled by the remaining three letters: E, L, and H.
• The number of ways to arrange 3 distinct items in 3 positions is calculated as 3! (3 factorial).
• 3! = 3 × 2 × 1 = 6.
• Therefore, there are exactly 6 different words that satisfy the given conditions.
Why the other options are wrong
- (a) 24
- 24 is the total number of permutations of all 5 letters (5!), ignoring the fixed start and end constraints.
- (b) 18
- 18 is not a valid permutation result for this specific constraint set and does not correspond to any standard calculation for this problem.
- (c) 12
- 12 would imply arranging 4 items (4!/2 or similar error), but only 3 letters remain to be arranged in the middle.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2020, held on 4 October 2020. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.