Posted by Joel on November 06, 2002 at 23:22:11:
Q: Out of a group of 7 women and 9 men, how many ways are there to select a committee of 5 people if the committee must include at least 1 woman?
DEFINITELY the answer is: the number of ways to select any committee of 5 out of 16 people, minus the number of ways to select a committee with NO women, so
C(16,5) - C(9,5) = 16!/5!11! - 9!/5!4! = 4242
But, why don't I get the same answer if I say: number of ways to select 1 women out of 7 times number of ways to select any 4 people out of the remaining 15:
C(7,1) * C(15,4) = 7 * 15!/4!11! = 9555 clearly wrong, but who's getting double-counted here?