• # question_answer 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 colours is [EAMCET 1991] A) $(2n+2)\ !$ B) $(2n+2)\ !\ \times 2$ C) $(n+1)\ !\ \times 2$ D) $2{{\{(n+1)\ !\}}^{2}}$

Since the balls are to be arranged in a row so that the adjacent balls are of different colours, therefore we can 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. Now $(n+2)$ places are created between $n+1$ white balls which can be filled by $(n+1)$ black balls in $(n+1)\ !$ ways. So the total number of arrangements in which adjacent balls are of different colours and first ball is a white ball is$(n+1)\ !\ \times (n+1)\ !\ ={{[(n+1)\ !]}^{2}}$. But we can begin with a black ball also. Hence the required number of arrangements is$2{{[(n+1)\ !]}^{2}}$.