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

Since the committee must have between 2 and 5 men, an alternate method
(which involves more calculation) is to take
$\displaystyle\binom{10}{2}\binom{12}{4}+\binom{10}{3}\binom{12}{3}+\binom{10}{4}\binom{12}{2}+\binom{10}{5}\binom{12}{1}=65,559$.
