Question

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?

• A. $47, 73$
• B. $48, 74$
• C. $48, 72$
• D. $49, 72$

Answer 1

Answer: (C)

$48, 72$

Answer 2

(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.