There are (n + 1) white and (n + 1) black bal | vedclass

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(D) To arrange $(n + 1)$ white balls and $(n + 1)$ black balls such that no two adjacent balls are of the same colour,the balls must alternate in colo

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$2 \{(n + 1)!\}^2$

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(D) To arrange $(n + 1)$ white balls and $(n + 1)$ black balls such that no two adjacent balls are of the same colour,the balls must alternate in colour.Case $1$: The arrangement starts with a white ball.The sequence must be $W, B, W, B, \dots, W, B$.The $(n + 1)$ white balls can be arranged in $(n + 1)!$ ways,and the $(n + 1)$ black balls can be arranged in $(n + 1)!$ ways.Total ways for this case $= (n + 1)! 	imes (n + 1)! = \{(n + 1)!\}^2$.Case $2$: The arrangement starts with a black ball.The sequence must be $B, W, B, W, \dots, B, W$.Similarly,the total ways for this case $= (n + 1)! 	imes (n + 1)! = \{(n + 1)!\}^2$.Since these two cases are mutually exclusive,the total number of arrangements is $\{(n + 1)!\}^2 + \{(n + 1)!\}^2 = 2 \{(n + 1)!\}^2$.

There are $(n + 1)$ white and $(n + 1)$ black balls,with each set numbered $1$ to $n + 1$. The number of ways in which the balls can be arranged in a row so that no two adjacent balls are of the same colour is

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