A) \[(2n+2)\ !\]
B) \[(2n+2)\ !\ \times 2\]
C) \[(n+1)\ !\ \times 2\]
D) \[2{{\{(n+1)\ !\}}^{2}}\]
Correct Answer: D
Solution :
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\[2{{[(n+1)\ !]}^{2}}\].
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