Question 1086412

1.  If the committee has exactly 3 women:

#ways of selecting 3 women * # ways of selecting 6-3=3 men:
(6C3) x (4C3) = 6!/(3!*3!) * 4!/(3!*1!) = 20 * 4 = 80  ways 


2.  If the committee has exactly 2 women:
(6C2) x (4C4) = 6!/(4!*2!) * 4!/(4!0!) = 15 * 1 = 15  ways

Those are the only two possibilities for #women and #men because if you tried to form the committee with just 1 woman, there wouldn't be enough men to make a committee of 6.

Thus, the answer is  80+15 =  {{{ highlight(95) }}} ways