In how many ways can 5 children be arranged in a line such that (i) t

1.	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.

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.