Poker hand probabilities

Homework Statement

A standard deck of playing cards has all the red (heart, diamond) cards removed. Find the probability of obtaining a 2 pair when 5 cards are dealt.

Homework Equations

The combination formula:

$\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}$

The Attempt at a Solution

The way I understood it was that there were 13 pairs of cards. In order to get a two pair, I need to choose 2 pairs from the 13, and choose 2 cards from each pair. For the fifth card, I just pick any pair from the remaining 11, and choose 1 of the pair. Then, that number of combinations I divide by the total number of ways I can pick 5 cards from 26. This is what I came up with:

$P(2 pair) = \dfrac{\dbinom{13}{1} \dbinom{12}{1} \dbinom{11}{1} (\dbinom{2}{2})^2 \dbinom{2}{1}}{\dbinom{26}{5}}$

However, the actual answer is slightly different:

$P(2 pair) = \dfrac{\dbinom{13}{2} \dbinom{11}{1} (\dbinom{2}{2})^2 \dbinom{2}{1}}{\dbinom{26}{5}}$

I can see why the actual answer makes sense, but I can't see why the answer I came up with does not. To me, they seem to be the exact same expression. Am I misunderstanding the meaning behind a combination here? Am I misunderstanding the meaning behind multiplying the number of combinations?

Some light shed on this problem would be much appreciated. Thank you.