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