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

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(D) Since the balls are to be arranged in a row such that adjacent balls are of different colours,we can start the arrangement with either a white bal

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

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(D) Since the balls are to be arranged in a row such that adjacent balls are of different colours,we can start the arrangement with either a white ball or a black ball.Case $1$: If we start with a white ball,the $(n + 1)$ white balls can be arranged in $(n + 1)!$ ways. This creates $(n + 2)$ possible positions (including ends) for the black balls. However,to ensure alternating colours,the $(n + 1)$ black balls must occupy the $(n + 1)$ specific gaps between the white balls. These $(n + 1)$ black balls can be arranged in these gaps in $(n + 1)!$ ways.Thus,the total arrangements starting with a white ball is $(n + 1)! 	imes (n + 1)! = \{(n + 1)!\}^2$.Case $2$: Similarly,if we start with a black ball,the total arrangements will be $(n + 1)! 	imes (n + 1)! = \{(n + 1)!\}^2$.Total arrangements = $\{(n + 1)!\}^2 + \{(n + 1)!\}^2 = 2\{(n + 1)!\}^2$.

There are $(n + 1)$ white and $(n + 1)$ black balls,each set numbered $1$ to $(n + 1)$. The number of ways in which the balls can be arranged in a row so that the adjacent balls are of different colours is

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