UPSC Prelims 2020 · CSAT
Question 67: Basic Numeracy
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?
AnswerUPSC’s answer is (d).
D must be first and I last, so the three letters E, L and H fill the middle three places in any order. The number of arrangements is 3! = 3 × 2 × 1 = 6. Remember: fix the constrained positions first and permute only the free letters.
How you could have got this
No shortcut here: in a “how many” question, ruling out options doesn’t help. Every statement has to be judged.
Which options go is worked out from UPSC’s key and the options, not by a model.
Why each option is right or wrong
- (a) 24 is 4!, which would leave only one end letter fixed.
- (b) 18 is not a factorial count here.
- (c) 12 does not come from 3 free letters.
- (d) Right: 3! = 6.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2020, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.