This is a really short question which I am sure is basic to most people on this board. In my class we have a combination question which I don't understand the solution to.
Question asks to find the number of ways you can make two teams of mixed gender (male, female is a team) for a game of tennis. There are $4$ males and $3$ females in the population.
The solution the teacher says is $2 \times {4 \choose 2}\times {3 \choose 2}$
I don't get this because how does the above formula ensure that you don't get 2 males on the same team or two females on the same team? How does the above ensure each team has a male and female?
Best Answer
For making $2$ mixed teams you'll need $2$ females and $2$ males.
You have $4\choose 2$ ways to choose the $2$ males and $3\choose 2$ ways to choose the $2$ females.
Now you can pair them in two different ways since each female can go with either male.
So that's $2\times {4\choose 2} \times {3\choose 2}$
EDIT for better understanding:
Note that $2\times {4\choose 2} \times {3\choose 2}=36$
Now, let's solve the problem somehow different.
There are teams A and B on each tennis game.
For team A we've got $4$ males and $3$ females, so $4\times 3=12$ possible A teams.
For team B we've got $3$ remaining males and $2$ remaining females, so $3\times 2=6$ possible B teams.
So we have $12\times 6=72$ ways to create a A & B team. But taking into account that team A & B are indistinguishable, we have to divide it by $2$ (Albert & Berta vs Charles & Diana is the same that Charles & Diana vs Albert & Berta). And $\frac{72}{2}=36$. Calculated differently, but as we see is the very same number.