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(C) To arrange $n$ white and $n$ black balls such that no two balls of the same colour are adjacent,the arrangement must be alternating.There are two

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$2(n!)^2$

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(C) To arrange $n$ white and $n$ black balls such that no two balls of the same colour are adjacent,the arrangement must be alternating.There are two possible patterns for the sequence of colours:$1$. $W, B, W, B, ..., W, B$$2$. $B, W, B, W, ..., B, W$For the first pattern,the $n$ white balls can be arranged in $n!$ ways and the $n$ black balls can be arranged in $n!$ ways. Thus,there are $n! 	imes n! = (n!)^2$ ways.For the second pattern,similarly,there are $n! 	imes n! = (n!)^2$ ways.Total number of ways = $(n!)^2 + (n!)^2 = 2(n!)^2$.

There are $n$ white and $n$ black balls marked $1, 2, 3, ..., n$. The number of ways in which we can arrange these balls in a row so that neighbouring balls are of different colours is

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