UPSC Prelims 2022 · CSAT
Question 24: Basic Numeracy
The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?
AnswerUPSC’s answer is (c).
Treat the letters as a row of five places. A and E can occupy places (1, 4) or (2, 5), and in either order: 2 × 2 = 4 ways. The other three letters fill the remaining three places in 3! = 6 ways. Total 4 × 6 = 24 arrangements. The gap rule (exactly two letters between) fixes only the places of A and E, so count those first and then permute the rest.
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) 12 forgets that A and E can swap places.
- (b) 18 does not result from this count.
- (c) Correct: 4 ways for A and E times 6 for the rest = 24.
- (d) 36 would assume three possible gaps, but a row of five gives only two.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2022, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.