Question

A committee of $6$ is chosen from $10$ men and $7$ women so as to contain atleast $3$ men and $2$ women. If $2$ particular women refuse to serve on the same committee, the number of ways of forming the committee is

• A. 7800
• B. 8610
• C. 810
• D. 7700

Answer 1

The correct option is A 7800

We have to exclude the cases when $2$ particular women will be together from all cases that could make

${}^{10}{C}_3$ & ${}^7{C}_3$ $-3$ men and $3$ women selected from group.

${}^{10}{C}_4$ & ${}^7{C}_2$ $-4$ men and $2$ women selection
When two particular women will be together:

${}^{10}{C}_4$ & ${}^5{C}_0$ $-4$ men and $0$ women selection

${}^{10}{C}_3$ & ${}^5{C}_1$ $-3$ men and $1$ women selection

Final Solution-

(${}^{10}{C}_3{.}^7{C}_3+{}^{10}{C}_4{.}^7{C}_2$) $-$ (${}^{10}{C}_4{.}^5{C}_0+{}^{10}{C}_3{.}^5{C}_1$)

$8610-810$ $=$ $7800$