A committee of 12 is to be formed from 9 wome | vedclass

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(D) Total committee size is $12$. Let $W$ be the number of women and $M$ be the number of men. We require $W + M = 12$ and $W \ge 5$.$(i)$ Women are i

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$2702, 1008$

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(D) Total committee size is $12$. Let $W$ be the number of women and $M$ be the number of men. We require $W + M = 12$ and $W \ge 5$.$(i)$ Women are in the majority if $W > M$. Since $W + M = 12$,this implies $W > 6$. Possible values for $W$ are $7, 8, 9$.Number of ways = $^9C_7 	imes ^8C_5 + ^9C_8 	imes ^8C_4 + ^9C_9 	imes ^8C_3$$= (36 	imes 56) + (9 	imes 70) + (1 	imes 56) = 2016 + 630 + 56 = 2702$.(ii) Men are in the majority if $M > W$. Since $W + M = 12$,this implies $M > 6$. Possible values for $W$ are $5, 6$ (since $W \ge 5$).Number of ways = $^9C_5 	imes ^8C_7 + ^9C_6 	imes ^8C_6$$= (126 	imes 8) + (84 	imes 28) = 1008 + 2352 = 3360$.Wait,re-evaluating: The question asks for committees where $W \ge 5$. If $M > W$,then $W$ can be $5$ or $6$.For $W=5, M=7$: $^9C_5 	imes ^8C_7 = 126 	imes 8 = 1008$.For $W=6, M=6$: This is not a majority. Thus,only $W=5$ gives a male majority.Correct answer is $2702, 1008$.

$A$ committee of $12$ is to be formed from $9$ women and $8$ men such that at least $5$ women are included. Find the number of committees in which women are in the majority and the number of committees in which men are in the majority,respectively.

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