The Probability of a two pair in poker 4

Question

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)?

Answer 1

You pick the two (distinct) ranks that are to occur as pairs. That's $13\choose 2$.

You pick the suits for the higher of these two ranks. That's $4\choose 2$.

You pick the suits for the lower of these two ranks. That's $4\choose 2$.

You pick the rank of the additional card from the ranks not used for the pairs. That's $11\choose 1$.

You pick the suit for that card. That's $4\choose 1$.