UPSC Prelims 2004 · General Studies
Question 139: Basic Numeracy
In how many different ways can six players be arranged in a line such that two of them, Ajit and Mukherjee, are never together?
AnswerThe answer is (d).
Six players can stand in a line in 6! = 720 ways. If Ajit and Mukherjee stand together, treat them as one block: 5! x 2 = 240 arrangements. So the arrangements with them never together are 720 - 240 = 480. Remember the method: total, minus the cases where they are together.
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.
This follows an unofficial answer: UPSC has not published its key for this year, and if that answer is wrong, so is this path.
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) 120 is 5!, the block arrangements without the swap.
- (b) 240 is the number with the pair together.
- (c) 360 is half of 720, not the count asked.
- (d) Correct: 720 - 240 = 480.
Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2004, General Studies. UPSC's own paper isn't online; the words are from a published compilation. The paper on upsc.gov.in.