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,
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) two of them, Rajan and Tanvy, are always together

Consider Rajan and Tanvy as a group which can be arranged in 2! = 2 ways. The 3 children with this 1 group can be arranged in 4! = 24 ways.

The total number of possibilities in which they both come together is 2×24 = 48 ways.

(ii) Two of them, Rajan and Tanvy, are never together

Two of them are never together = total number of possible ways of sitting – total number of ways in which they sit together.

A total number of possible way of arrangement of 5 students is 5! = 120 ways.

Therefore, the total number of arrangement when they both don’t sit together is = 120 – 48 = 72.