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Posted by MathBard on November 07, 2002 at 00:00:31:

In Reply to: selecting a committee posted by Joel on November 06, 2002 at 23:22:11:

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!
MB

: 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?

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