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

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(C) To arrange $n$ white and $n$ black balls in a row such that no two balls of the same colour are adjacent,the balls must alternate in colour.There

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$2(n!)^2$

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(C) To arrange $n$ white and $n$ black balls in a row such that no two balls of the same colour are adjacent,the balls must alternate in colour.There are two possible patterns for this arrangement:Case $1$: The sequence starts with a white ball: $W_1, B_1, W_2, B_2, ..., W_n, B_n$.In this case,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.Case $2$: The sequence starts with a black ball: $B_1, W_1, B_2, W_2, ..., B_n, W_n$.Similarly,there are $n! 	imes n! = (n!)^2$ ways for this case.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 are

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