How Many Ways to Choose Dance Partners if Choosing Both Men and Women? I was working through supplementary exercises in Grimaldi, and I came across a question about dance partners--it asked how to choose a clean-up team of 8 from a group of 12 men, and then asked how many different ways to pair up 8 women from those 12 men.  The answer was simple enough: find the number of permutations of size 8 from a set of 12 men (i.e., $\frac{12!}{4!})$.  But that got me wondering, if we want to choose both the men and the women from a set of 12 each, how does the answer change?  Since I'm completely new to self-studying math, I thought I'd pose the question to see if my approach to answering the question is valid (I thought it might still have something to do with permutations, but I wasn't sure).
So let's say there are 12 men and 12 women at a dance.  How many ways are there to pair off 8 men and 8 women as dance partners?
 A: Here's my guess.  First we choose 8 men from the 12 total men, which is $\binom{12}{8}$.  Then, once we have the 8 men chosen, we find the permutations of 8 women from the set of 12, which is $\frac{12!}{(12-8)!} = \frac{12!}{4!}$.  The total number of ways should therefore be:
$$
\binom{12}{8} \times \frac{12!}{4!}
$$
A: There are $\binom{12}8$ ways to choose the men, and $\binom{12}8$ ways to choose the women.  Once the people have been chosen there are $8!$ ways to pair them off.  (Stand the women in a line.  There are $8!$ ways to shuffle the men, then we pair the first man with the first woman, the second man with the second woman, and so on.)
The answer is $$\binom{12}{8}^2\cdot8!=9,879,408,000
$$
A: Let us simplify the question.  Imagine you have to choose 2 pairs from 3 pairs of men  and woman (M1-M3,W1-W3) to form partners.
Step 1:
Chose 2 from  3 men,  $\binom{3}{2}$ =3 choices.
Step 2:
Chose 2 from  3 women, $\binom{3}{2}$ =3 choices.
Step 3:
Then to pair them $\to$
If no need  to line them up, 3x3 =  9

If need line them up, 3x3x2! =18

Now, we can answer the  original question.
