Discrete Math Boys and Girls Problem 4: Boys and Girls Consider a set of m boys and n girls. A group is called homogeneous if it consists of all boys or all girls. In the following questions, practice the multiplication and the addition rule, and state exactly where you have used them.
(a)How many teams of two can we make?
(b)How many teams of three can we make?
(c)How many non-homogenous teams of two can we make?
(d)How many non-homogenous teams of three can we make?
This is what I got and idk if its right.
a. We can make n(n-1) or m(m-1) teams of two. (Multiplication Rule)
b. We can make n(n-1)(n-2) or m(m-1)(m-2) teams of three. (Multiplication Rule)
c. We can make (n+m) non-homogenous teams of two. (Addition rule)
 A: You’re a bit off track, I’m afraid. I’ll discuss the case of two-person teams in some detail; see if you can then apply the ideas to correct your answers for the case of three-person teams.
Your $n(n-1)$ is the number of ways to choose an ordered team of two girls. There are two problems with this: we don’t want to count the team of Farah-and-Layla as different from the team of Layla-and-Farah, and we want to count all two-person teams, not just those formed of girls.
Your $n(n-1)$ counts each team of two girls twice: we can form the team consisting of Farah and Layla, for instance, by picking Farah and then Layla, or by picking Layla and then Farah. Thus, the actual number of teams consisting of two girls is $\frac12n(n-1)$. Similarly, there are $\frac12m(m-1)$ teams consisting of two boys. Altogether, then there are
$$\frac12n(n-1)+\frac12m(m-1)=\frac{n(n-1)+m(m-1)}2\tag{1}$$
homogeneous teams of two. The problem doesn’t explicitly ask for this, but we’ll need it later.
If we don’t care about homogeneity, we have a lot more ways to choose a team of two. The first member can be any one of the $m+n$ boys and girls, and the second can be any one of the $m+n-1$ remaining boys and girls. That gives us $(m+n)(m+n-1)$ ordered teams of two: once again we’ve counted each team twice, once for each of the two orders in which we could have picked it. Thus, there are
$$\frac{(m+n)(m+n-1)}2\tag{2}$$
teams of two altogether. To get the number of non-homogeneous teams of two we must subtract $(1)$ from $(2)$; I leave it to you to do so and to do the necessary algebra to simplify the result.
