In how many ways can 5 children be arranged in a line such that
(i) two of them, Rajan and Tanvy, are always together?
(ii) two of them, Rajan and Tanvy, are never together?

Answer

(i) 48 (ii) 72

Explanation

Total number of ways of arranging 5 children in a row = 5! = 120.
(i) Considering Rajan + Tanvy as one child, this one and 3 more can be arranged in 4! = 24 ways.
Rajan and Tanvy can be arranged among themselves in 2! = 2 ways.
Required number of ways `=(24 xx2)=48`.
(ii) When Rajan and Tanvy are never together, the number of arrangements `=(120-48)=72.`