How many ways can we get 2 a's and 2 b's from aabb? We have the following group: $aabb$

It is commutative, so abab is the same as aabb.

I have figured out this is a combinatorics question.  Because abab is the same as aabb.  I was how to solve these problems with the blank slot method, i.e. _ _ _ _.  

If I do this manually, it's clear to me the answer is 6,

aabb
abab
abba
baba
bbaa
baab

Which is the same as $$\binom{4}{2}$$

But I don't really understand why this is true? How is this supposed to be done without brute forcing the question?
You have four slots. Choose two of them to be a, and the other two will be forced to be b. There are $\binom 4 2$ ways to choose two slots from four, giving the answer.
This is a permutation of $4$ objects with $2$ groups of $2$ identical elements. Answer is therefore $\frac{4!}{(2! \times 2!)}$

Or, this can be also the problem of selecting $2$ places out of $4$ to replace $b$ with $a$ in the string $bbbb$, in which case the answer is $\binom{4}{2}$