Ways a committee is selectd with at least 2 men and 1 woman.

For this question:

A committee of six is to be selected from a group of ten men and 12 women. In how many ways can the committee can be chosen if it has to contain at least two men and one woman?

Working I have attempted so far:

1.)$$\binom{22}{6}-\left(\binom{10}{6}+\binom{10}{5}\binom{12}{1}+\binom{12}{6}\right)=70959$$

So basically I am not really sure if it is right, It seems right to me though. Please help in correcting my working if wrong

• Your work is correct since you subtracted the number of committees that were not allowed from the total possible ways of selecting the committees. – N. F. Taussig Apr 10 '15 at 12:34
• It should be $\binom {12}5\binom{10}1$ instead of $\binom {10}5\binom{12}1$. – Hypergeometricx Apr 10 '15 at 16:54

you want to have at least 2 men and 1 woman in the committee, so you have to subtract the ways that all of the committee are men, all of the committee are women and also the committee has just 1 man and 5 women. the answer is: $\\$

$$\binom{22}{6}-\left(\binom{10}{6}+\binom{10}{1}\binom{12}{5}+\binom{12}{6}\right)$$

• ahhh thanks for explanation guys, did not realize my mistake – Itakura Apr 11 '15 at 11:14

Let (m, w), where m = no. of men forming the committee and w = no. of women forming the committee, denote the number of committee formed.

Choosing at least 2 men and 1 woman = C(22, 6) - (0,6) - (1,5) - (6,0)

Therefore the number of ways = C(22, 6) - C(12, 6) - C(10, 1)*C(12, 5) - C(10, 6) = 65559.

Since the committee must have between 2 and 5 men, an alternate method

(which involves more calculation) is to take

$\displaystyle\binom{10}{2}\binom{12}{4}+\binom{10}{3}\binom{12}{3}+\binom{10}{4}\binom{12}{2}+\binom{10}{5}\binom{12}{1}=65,559$.