# How many ways can a committtee of $2$ women and $3$ men be selected from a group of $5$ women and $7$ men if two of the men refuse to serve together?

I am trying to learn permutations and combinations. Please help me to solve the second part of the question.

From a group of $5$ women and $7$ men, how many different committees of $2$ women and $3$ men can be formed? What if $2$ of the men are feuding and refuse to serve on the committee together?

I gather that you understand that the number of committees of two women and three men that can be selected from five women and seven men is $$\binom{5}{2}\binom{7}{3}$$ Below is a hint for the second part of the question.

Hint: Consider three groups of people: the five women, the five men who are not feuding, and the two men who are feuding. There are two ways to form the committee so that the two men who are feuding are not both selected:

1. Neither of the feuding men is selected. Choose two of the five women and three of the five men who are not feuding.
2. Exactly one of the feuding men is selected. Choose two of the five women, two of the five men who are not feuding, and one of the two men who are feuding.

A committee with 2 women and 3 men is a group of 2 women together with a group of 3 men.

How many groups of 2 women are there? You have 5 ways to pick the first woman, 4 ways to pick the second, and then you have to divide by 2, because picking Alice and then Beatrice is the same as picking Beatrice and then Alice. Order does not matter. Thus you have

$$\frac{5\cdot4}{2} = {5\choose2}$$

Similarly for the group of 3 men. You have $7\cdot6\cdot5$ ways of picking them, but then you have to discount the different orders that pick the same groups. Picking the men in these orders form always the same group:

$$ABC\\ ACB\\ BAC\\ BCA\\ CAB\\ CBA$$

So 6 different orders produce the same group, so every 6 different orders give the same group. Why 6? Because 6, $3!$, is the number of ways you can order 3 elements. Thus you have

$$\frac{7\cdot6\cdot5}{3\cdot2} = {7\choose3}$$

You now multiply both together and get

$${5\choose2}\cdot{7\choose3}$$

diferent committees.

In the general case, if you have a group of $n$ people and you want to form a group with $k$ of them, you get

$${n\choose k }$$

total groups, if there are no further restrictions. Notice that this formula matches our results for the group of 2 women and for the group of 3 men.

For the second part, let us call Bob and John to the two guys refusing to work together. The number of valid committees is the number of committees with none of them plus the number of committees with Bob and 2 guys other than John plus the number of committees with John and two guys other than Bob. Can you try and calculate that?