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)

(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.