There are n white and n black balls marked 1, | vedclass

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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 balls must alternate in colour.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 balls must alternate in colour.There are two possible patterns for the arrangement:$1$. $B, W, B, W, \ldots, B, W$ (starting with Black)$2$. $W, B, W, B, \ldots, W, B$ (starting with White)For each pattern, the $n$ black balls can be arranged in $n!$ ways and the $n$ white balls can be arranged in $n!$ ways.Thus, the number of ways for the first pattern is $n! 	imes n! = (n!)^2$.Similarly, the number of ways for the second pattern is $n! 	imes n! = (n!)^2$.Total number of ways = $(n!)^2 + (n!)^2 = 2(n!)^2$.

There are $n$ white and $n$ black balls marked $1, 2, 3, \ldots, n$. 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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