so just a quick note. My question is very similar (but not the same) to the question at this link:
Probability of getting two pair in poker
I calculated the probability of getting two pair as
$P=$ $$\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$
This was wrong, and the correct answer is:
$P=$ $$\frac{{13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$
Upon looking at the correct answer, I now understand that:
$$\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose2}}{52\choose5}$$
Introduces redundancy for the two pair. Ie it counts both a (6H, 6D, 3H, 3D, X) and a (3H, 3D, 6H, 6D, X). Which is why the correct usage is $13\choose2$, but there is another thing I don't understand.
My problem is I feel like by the same way my original answer introduces redundancy, the correct answer also has a redundancy with the remaining card. How does
$${13\choose2}{4\choose2}^2{11\choose1}{4\choose1}$$
not count (6H, 6D, 3H, 3D, X) and (X, 6H, 6D, 3H, 3D)?