The number of ways of choosing a committee fr | vedclass

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(C) Let $m$ be the number of men and $w$ be the number of women in the committee.We are given that $w = 2m$ and $m \ge 2$.Since there are $4$ men and

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$94$

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(C) Let $m$ be the number of men and $w$ be the number of women in the committee.We are given that $w = 2m$ and $m \ge 2$.Since there are $4$ men and $6$ women available,we have the constraints $m \le 4$ and $w \le 6$.Substituting $w = 2m$ into $w \le 6$,we get $2m \le 6$,which implies $m \le 3$.Thus,the possible values for $m$ are $2$ and $3$.Case $1$: If $m = 2$,then $w = 2(2) = 4$. The number of ways is $^4C_2 	imes ^6C_4 = 6 	imes 15 = 90$.Case $2$: If $m = 3$,then $w = 2(3) = 6$. The number of ways is $^4C_3 	imes ^6C_6 = 4 	imes 1 = 4$.Total number of ways = $90 + 4 = 94$.

The number of ways of choosing a committee from $4$ men and $6$ women so that the committee includes at least $2$ men and exactly twice as many women as men is

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