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

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(B) The total number of balls is $2n$. We draw $n$ pairs of balls one by one. For the first pair,the number of ways to choose one white and one black

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The total number of balls is $2n$. We draw $n$ pairs of balls one by one. For the first pair,the number of ways to choose one white and one black ball is $\binom{n}{1} 	imes \binom{n}{1} = n^2$. For the second pair,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 $(n^2) 	imes (n-1)^2 	imes \dots 	imes 1^2 = (n!)^2$. Given that $(n!)^2 = 14400$. Taking the square root on both sides,$n! = \sqrt{14400} = 120$. Since $5! = 120$,we have $n = 5$.

$A$ 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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