Let m denote the number of ways in which 5 bo | vedclass

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(D) For $m$: The boys and girls can be arranged in a line alternately in two ways: ($B$ $G$ $B$ $G$ $B$ $G$ $B$ $G$ $B$ $G$) or ($G$ $B$ $G$ $B$ $G$ $

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$10$

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(D) For $m$: The boys and girls can be arranged in a line alternately in two ways: ($B$ $G$ $B$ $G$ $B$ $G$ $B$ $G$ $B$ $G$) or ($G$ $B$ $G$ $B$ $G$ $B$ $G$ $B$ $G$ $B$). In each case,$5$ boys can be arranged in $5!$ ways and $5$ girls can be arranged in $5!$ ways. So,$m = 2 	imes 5! 	imes 5! = 2 	imes 120 	imes 120 = 28800$. For $n$: To arrange $5$ boys and $5$ girls in a circle such that no two boys are together,we first arrange the $5$ girls in a circle in $(5-1)! = 4! = 24$ ways. This creates $5$ gaps between them. The $5$ boys can be arranged in these $5$ gaps in $5! = 120$ ways. So,$n = 4! 	imes 5! = 24 	imes 120 = 2880$. Given $m = kn$,we have $28800 = k 	imes 2880$,which gives $k = 10$.

Let $m$ denote the number of ways in which $5$ boys and $5$ girls can be arranged in a line alternately,and $n$ denote the number of ways in which $5$ boys and $5$ girls can be arranged in a circle so that no two boys are together. If $m = kn$,then the value of $k$ is

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