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?
Best Answer
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:
$$ \binom{12}{8} \times \frac{12!}{4!} $$