From a group of 10 men and 8 women,the number | vedclass

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(A) We need to form a committee of $8$ members from $10$ men and $8$ women such that there are at most $5$ men and at least $5$ women. Let $M$ be the

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$8061$

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(A) We need to form a committee of $8$ members from $10$ men and $8$ women such that there are at most $5$ men and at least $5$ women. Let $M$ be the number of men and $W$ be the number of women. Since $M+W=8$,$M \le 5$ and $W \ge 5$. The possible cases are:Case $I$: $3$ men and $5$ women. Number of ways $= {}^{10}C_3 	imes {}^{8}C_5 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} 	imes \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 120 	imes 56 = 6720$.Case $II$: $2$ men and $6$ women. Number of ways $= {}^{10}C_2 	imes {}^{8}C_6 = \frac{10 	imes 9}{2 	imes 1} 	imes \frac{8 	imes 7}{2 	imes 1} = 45 	imes 28 = 1260$.Case $III$: $1$ man and $7$ women. Number of ways $= {}^{10}C_1 	imes {}^{8}C_7 = 10 	imes 8 = 80$.Case $IV$: $0$ men and $8$ women. Number of ways $= {}^{10}C_0 	imes {}^{8}C_8 = 1 	imes 1 = 1$.Total number of ways $= 6720 + 1260 + 80 + 1 = 8061$.

From a group of $10$ men and $8$ women,the number of ways of forming a committee of $8$ members with not more than $5$ men and not less than $5$ women is

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