UPSC Prelims 2010 · General Studies
Question 117: Basic Numeracy
In how many ways can four children be made to stand in a line such that two of them, A and B are always together ?
AnswerThe answer is (b).
Treat A and B as one block that always stays together. The block and the other two children make 3 units, which can be arranged in 3! = 6 ways. Inside the block, A and B can be in either order, 2 ways. Total = 6 x 2 = 12. Remember: for 'always together', glue the items, count arrangements of the units, then multiply by the orders inside the block.
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) 6 forgets the two orders of A and B inside the block.
- (b) Right: 3! x 2! = 12.
- (c) 18 has no basis in the count.
- (d) 24 is the total with no restriction (4!).
Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2010, General Studies, Series A. The paper on upsc.gov.in.