There are (n + 1) white and (n + 1) black balls, each set numbered 1 to n + 1. The number of ways in which the balls can be arranged in a row so that the adjacent balls are of different colors is

Since the balls are to be arranged in a row so that the adjacent balls are of different colours, we can therefore begin with a white ball or a black ball. If we begin with a white ball, we find that n + 1 white balls numbered 1 to n + 1 can be arranged in a row in(n + 1)! ways.

      $\times \mathrm{w}\times \mathrm{w}\times \mathrm{w}\times \mathrm{w}\times \dots \times \mathrm{w}$

Since white and black balls are alternate, we have exactly (n + 1) place for black balls marked with 'X'

So, the total number of arrangements in which adjacent balls are of different colours and first ball is a white ball is $(\mathrm{n}+1)!\times (\mathrm{n}+1)!=\left[\right(\mathrm{n}+1)!{]}^2$. But we can begin with a black ball also. Hence, the required number of arrangements is 2[(n + 1)!]².