How many different ways can 2 groups of 10 be divided into 10 groups of 2? You'll have to forgive me. Math is definitely not my specialty, but my curiosity got the best of me.
My wife watches the show "Are You The One". Doesn't interest me in the least, but it made me wonder about the odds. 
The premise, to my understanding, is this:
20 people--10 male and 10 female have to collectively figure out who their 'perfect matches' are. 
Each person has one other person who is supposed to be their perfect match. 
I was just wondering what the odds are, or how many combinations can be made of groups of 2 from 2 groups of 10.
At first I thought it was 10 factorial (3628800), but then that didn't seem quite right. Question one: Am I correct in assuming this is the number of permutations of 2 groups of 10 divided into 10 groups of 2 with one from each group?
Then, I thought that it was actually 10 * 10 + 9 * 9 + 8 * 8 ... + 1 * 1, or 385.
Question 2: Is this the number of  combinations of 2 groups of 10 divided into 10 groups of 2, and if so, what is this mathematical operation called (as opposed to factorial)?
Thanks in advance for the help!
 A: This answer is assuming that men can only be paired with women and vice-versa.
Line the men in a line side-by-side and do the same for the women in such a way that the men and women are facing one another and that each couple facing one another is considered a match. Notice that if we leave the men fixed, we can achieve any combination of matchings by simply re-ordering the women of which there are $10!$ ways of doing so in which case your original guess is correct.
A: You were close with your guess of $10!$. Assuming anyone can match with anyone else, then the answer is $19!!$, i.e., the double factorial of $19$. This is defined to be:
$$19!! = 19 \cdot 17 \cdot 15 \cdots 3 \cdot 1 = 654,729,075.$$
The reasoning is as follows: let's order the people alphabetically. The first person has a choice of $19$ people; now remove that pair, and $18$ people remain. Repeat the process: take the first person from those $18$, and he or she has a choice of $17$ partners. Remove that pair, and so on.
Note that the value above is also the product of $10$ numbers, just like $10!$, but is substantially larger than $10!$. Since each term in $19!!$ is roughly twice as big as the corresponding term in $10!$, the result is (very roughly) on the order of $2^{10}$ times as big. (It turns out to be $180$ times as big.)

Edit: It wasn't specified in the problem, but if the goal is to match men to women, then yes, your calculation of $10!$ is correct. The first woman has a choice of $10$ men, the next woman has a choice of $9$ men, and so on.
