There are (n + 1) white balls and (n + 1) bla | vedclass

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(D) To arrange the balls such that no two balls of the same color are adjacent,we must alternate the colors.There are two possible starting colors: wh

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

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(D) To arrange the balls such that no two balls of the same color are adjacent,we must alternate the colors.There are two possible starting colors: white or black.Case $1$: Starting with a white ball.We arrange the $(n + 1)$ white balls in $(n + 1)!$ ways.There are $(n + 1)$ spaces created by these white balls (including ends),but since we have $(n + 1)$ black balls to place in the gaps between the white balls,there is only one way to place them such that they alternate: $W B W B ... W B$.Thus,the number of ways to arrange them starting with white is $(n + 1)! 	imes (n + 1)! = {(n + 1)!}^2$.Case $2$: Starting with a black ball.Similarly,the number of ways to arrange them starting with black is $(n + 1)! 	imes (n + 1)! = {(n + 1)!}^2$.Total number of arrangements = ${(n + 1)!}^2 + {(n + 1)!}^2 = 2{(n + 1)!}^2$.

There are $(n + 1)$ white balls and $(n + 1)$ black balls. Each ball is numbered from $1$ to $(n + 1)$. In how many ways can these balls be arranged in a row such that no two balls of the same color are adjacent?

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