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.      ×w×w×w×w×…×wSince 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 (n+1)!×(n+1)!=[(n+1)!]2. But we can begin with a black ball also. Hence, the required number of arrangements is 2[(n + 1)!]2.