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 ?

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Q: 147 (IAS/2010)
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 ?

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,15,7,1,15,3,3

keywords: 

{'children': [0, 0, 1, 0], 'many ways': [2, 0, 2, 0], 'line': [4, 1, 3, 4]}

To solve the problem, we can consider A and B as a single unit. This gives us three units to arrange - AB, C, and D.

Now, we can arrange these three units in 3! ways = 6 ways.

Within the AB unit, there are two ways to arrange A and B. Therefore, the total number of ways to arrange the four children such that A and B are always together is 6 x 2 = 12.

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