Posted by Joel on November 07, 2002 at 00:11:06:
In Reply to: speculating at an answer... posted by MathBard on November 07, 2002 at 00:00:31:
: i think in your second formula you are NOT subtracting for committess of 2,3,4, or 5 women!Just those with ONE women. I dunno my probabiltiy or summations well enough to be sure! But I am sure if i bring thie to your attention you'll know if you are or not including them!
Thanks, but no, I still don't see what's wrong. It seemed legitimate to say: the number of ways to select 1 woman out of 7 -- obviously, 7 ways; now there are 15 people left, number of ways to select any 4 out of 15, which is 1365; so multiply 1365 by 7 giving 9555. Way too much, but I don't see where the logic fails.
: : 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?
Post a Followup