in how many ways can a committee of five be selected from 8 women and 5 men.

a) without restriction

b) if there must be three women on the committee

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- Jan 23rd 2011, 07:19 PMjohnsy123Comb application 2
in how many ways can a committee of five be selected from 8 women and 5 men.

a) without restriction

b) if there must be three women on the committee - Jan 23rd 2011, 07:29 PMProve It
a) There are 13 people to choose from and you must choose 5.

So the number of combinations of 5 people from 13 is $\displaystyle \displaystyle {13\choose{5}} = \frac{13!}{5!(13 - 5)!}$.

For part b) does that mean there must be EXACTLY three women or AT LEAST three women? - Jan 23rd 2011, 07:31 PMjohnsy123
never mind guys, i worked it out. :)

- Jan 23rd 2011, 07:37 PMProve It
You're welcome :P

- Jan 23rd 2011, 07:41 PMjohnsy123
Your working out is wrong, because you can't simply work out the number of combinations as there are independent variables, so what i did is i worked out the combinations of which you can variate 4 men and 8 women into a committee of 5, there were 5.

-If i knew how to use the symbols on MHF, i would show you what i done. - Jan 23rd 2011, 07:46 PMProve It
- Jan 24th 2011, 01:25 AMmr fantastic