# Thread: Committees with 12 men and 8 women

1. ## Committees with 12 men and 8 women

I think I'm beginning to understand this, so I would appreciate if someone could critique the following:

4. A group consists of 12 men and 8 women. How many ways are there to ...

4.a) form a committee of size 5?
$\left({20\atop 5}\right)$

4.b) form a committee of size 5 containing two men and three women?
$\left({12\atop 2}\right)*\left({8\atop 3}\right)$

4.c) form a committee of size 6 containing at least three women?

First approach: $\left({12\atop 3}\right)*\left({8\atop 3}\right)+\left({12\atop 2}\right)*\left({8\atop 4}\right)+\left({12\atop 1}\right)*\left({8\atop 5}\right)+\left({12\atop 0}\right)*\left({8\atop 6}\right)$

Second approach: $2^{20}-\left({12\atop 4}\right)*\left({8\atop 2}\right)-\left({12\atop 5}\right)*\left({8\atop 1}\right)-\left({12\atop 6}\right)*\left({8\atop 0}\right)$

2. Originally Posted by oldguynewstudent
I think I'm beginning to understand this, so I would appreciate if someone could critique the following:

4. A group consists of 12 men and 8 women. How many ways are there to ...

4.a) form a committee of size 5?
$\left({20\atop 5}\right)$

4.b) form a committee of size 5 containing two men and three women?
$\left({12\atop 2}\right)*\left({8\atop 3}\right)$

4.c) form a committee of size 6 containing at least three women?

First approach: $\left({12\atop 3}\right)*\left({8\atop 3}\right)+\left({12\atop 2}\right)*\left({8\atop 4}\right)+\left({12\atop 1}\right)*\left({8\atop 5}\right)+\left({12\atop 0}\right)*\left({8\atop 6}\right)$

Second approach: $12^{2}-\left({12\atop 4}\right)*\left({8\atop 2}\right)-\left({12\atop 5}\right)*\left({8\atop 1}\right)-\left({12\atop 6}\right)*\left({8\atop 0}\right)$
This should have the three subtracted from $\binom{20}{6}$ rather than $12^2$