In how many ways can four children be made to | vedclass

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(B) To ensure that $A$ and $B$ are always together,we treat them as a single unit or block $(AB)$.Now,we have this block $(AB)$ and the remaining two

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$12$

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(B) To ensure that $A$ and $B$ are always together,we treat them as a single unit or block $(AB)$.Now,we have this block $(AB)$ and the remaining two children,making a total of $3$ entities to arrange.The number of ways to arrange these $3$ entities in a line is $3! = 3 	imes 2 	imes 1 = 6$.Within the block $(AB)$,the two children $A$ and $B$ can arrange themselves in $2! = 2 	imes 1 = 2$ ways (i.e.,$AB$ or $BA$).Therefore,the total number of ways is $3! 	imes 2! = 6 	imes 2 = 12$.

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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