In how many ways can 5 children be arranged in a line such that

(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 $2 \times 24=48$ 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 $=120-$ $48=72$.