In how many ways can $5$ children be arranged in a line such that
(i) two particular children are always together
(ii) two particular children are never together?
Updated On: Jul 6, 2022
$ 47, 73$
$48, 74$
$48, 72$
$49, 72$
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The Correct Option isC
Solution and Explanation
(i) We consider the arrangements by taking $2$ particular children together as one and hence the $4$ children can be arranged in $4! = 24$ ways. Again two particular children taken together can be arranged in two ways. Therefore, there are $24 \times 2 = 48$ total ways of arrangement.
(ii) Among the $5! = 120$ permutations of $5$ children, there are $48$ in which two children are together. In the remaining $120 - 48 = 72$ permutations, two particular children are never together.