Question

A committee of 6 members is to be chosen from 10 women and 7 men such that it contains at least 3 women and at least 2 men. The number of different ways this can be done if 2 particular men do not agree to serve in the same committee is

• A. 8610
• B. 8710
• C. 7800
• D. None of these

Answer 1

The correct option is C 7800

Without considering the demand of the 2 men, the total number of committees
${=}^{10}{C}_3{.}^7{C}_3{+}^{10}{C}_4{.}^7{C}_2=8610$ (two cases: 3 women, 3 men, and 4 women, 2 men)
If we select the two particular men in the committee, the remaining $4$ members can be selected in
${}^{10}{C}_3{.}^5{C}_1{+}^{10}{C}_4{.}^5{C}_0=810$ (two cases: 3 women, 1 man, and 4 women, 0 men)
$\therefore$ The number of committees in which they work together $=810$
$\therefore$The number of committees in which they do not work together $=8610-810=7800$