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? 

 (a) 6 (b) 12 (c) 18 (d) 24


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(b) Take, A and B to be always together as a single entity.

Now, total no. of children = 4 – 2 + 1 = 3

These can be arranged in 3! ways and A, B can be arranged among themselves in 2! ways.

Hence, no. of arrangements such that A and B are always together = 3! × 2! = 3 × 2 × 2 = 12

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