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

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.
