Has this problem over-counted the possible combinations? Spending the weekend reviewing a few topics on the Pure Math 30 website. I have a question about this example.
If the order doesn't matter, why don't we divide by $2!$? Isn't just multiplying the two combinations together taking into account order?
I have not taken very advanced courses, so I would really appreciate an answer that isn't way too over my head!
Thank you in advance fellow math lovers!

 A: The fundamental counting principle says that if you have $k$ independent choices to make, where the first has $n_1$ options, the second $n_2$ options, and so on, then the total number of ways you can make all those choices is equal to $n_1n_2\cdots n_k$. There is no dividing by factorials going on here.
The time to divide by $m!$ is when you've made $m$ distinct choices in order, one after the other, all from the same pool of options, but you aren't supposed to take order into account as you were only trying to count the number of unordered sets of $m$ distinct choices in the first place, so you've overcounted by a factor of $m!$ (since the ordered selections can be grouped into permutations of unordered selections, and there are $m!$ different permutations of any given unordered selection).
Compare the following. Suppose there are $3$ boys and $5$ girls. Problem One: how many ways are there of choosing one boy and one girl? Answer: $3\times 5$. You don't divide by $2!$ (indeed you can't; that wouldn't be an integer) because the choice of boy and choice of girl are not drawn from the same pool of options. On the other hand, consider Problem Two: how many ways are there of choosing a pair of girls? If you pick girl one, that's $5$ possible choices, and then if you pick girl two there are $4$ possible choices, but since both girls were drawn from the same original pool of options we must divide by $2!$, obtaining $(5\times 4)/(2\times 1)=10$. You can even pick $3$ boy names and $5$ girl names and check by hand everything I've said.
