A bag contains n white and n black balls. Pai | vedclass

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(B) The total number of ways to draw $n$ pairs of balls such that each pair contains one white and one black ball is given by the product of the numbe

Vedclass · Accepted Answer

$5$

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(B) The total number of ways to draw $n$ pairs of balls such that each pair contains one white and one black ball is given by the product of the number of ways to choose one white and one black ball for each successive draw.For the first draw,there are $n$ white and $n$ black balls. The number of ways to pick one white and one black ball is $\binom{n}{1} 	imes \binom{n}{1} = n^2$.For the second draw,there are $(n-1)$ white and $(n-1)$ black balls remaining. The number of ways is $\binom{n-1}{1} 	imes \binom{n-1}{1} = (n-1)^2$.Continuing this until the last pair,the total number of ways is $(n 	imes (n-1) 	imes \dots 	imes 1)^2 = (n!)^2$.Given that $(n!)^2 = 14400$,we have $n! = \sqrt{14400} = 120$.Since $5! = 120$,we find $n = 5$.

$A$ bag contains $n$ white and $n$ black balls. Pairs of balls are drawn at random without replacement successively,until the bag is empty. If the number of ways in which each pair consists of one white and one black ball is $14400$,then $n$ is equal to

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