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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?

Masclins
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Kelbe
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1 Answers1

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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.

Masclins
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  • how do you make sure in your 4 choose 2 * 3 choose 2 that you are not counting the probability that 2 males are on same team? 4 choose 2 * 3 choose 2 also counts the number of ways you can count 2 males and 2 females but it doesn't guarantee the order.. – Kelbe May 12 '15 at 15:23
  • @Kelbe I'm counting the ways I can choose $2$ males and then $2$ females. I garantee they won't be in the same team because I only multiply it by $2$: For each female, only $2$ possible pairs, with the man or the woman. Otherwise, rather than by $2$ I'd multiply by $4\choose 2$ again, making all the possible couples out of the $4$ selected players. – Masclins May 12 '15 at 15:29
  • This doesn't make sense to me because I still don't get how you can guarantee the ORDER. I guess that you don't really care about the order and just want the probability of your desired selection....kind of confusion but atleast this will get me through the tests. – Kelbe May 12 '15 at 15:40