Hint 1: How many ways to choose a committee of $5$ from these $8$ people such that at least $3$ women are on the committee?
Hint 2: Of those, how many such will have a male president and a male vice president? (Recall that there must be at least three women!)
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If we have a committee of $4$ women, there are only two questions to answer: Which positions can we put the man in? Which man has that position? Well, either the man committee member is pres., vice pres., or secretary, so there are $3$ positions the man can end up in. Also, there are $4$ men who could take that position. Thus, there are $$3\cdot 4=12$$ such committees with only one man. Clearly, we can't have an all-woman committee (since there are only $4$ of them), and we are required to have at least $3$, so now we have to determine how many $3$-woman committees there are. It's still easier to work with the men, since there will be fewer of them, and again, we've got two questions to answer: Which positions can the two men can be placed distinctly in? How many ways can we pick the two men? There are $4$ men, of whom $2$ are chosen for the committee, so that there are $$\binom 4 2=6$$ ways to pick the men for the committee. Suppose we've picked them, and need only assign positions--let's call them Al and Bob, for distinction. We could have: both of them as secretaries; Al as pres. and Bob as a sec.; Al as vice pres. and Bob as a sec.; Al as pres. and Bob as vice pres.; Bob as pres. and Al as a sec.; Bob as vice pres. and Al as a sec.; Bob as pres. and Al as vice pres.--in total, that's $7$ different ways that the ranks can be divided among the two chosen men. Thus, there are $$6\cdot 7=42$$ such committees with only two men, and so there are $54$ committees with at least three women. However, some of those had females as secretaries only, which we don't want. Well, how many of those were there? It only happened with the $2$-man committee, and only when one was pres. and the other was vice pres. There were $6$ ways to pick the two men, and $2$ ways to rank them as pres. and vice pres., so that's $$6\cdot 2=12$$ committee configurations to discard, leaving $$54-12=42$$ committees of the type we wanted.