Calculating 2 Pair Probability in a Modified Deck of Cards

[walk_w/o_aim]

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.

 

[hgfalling]

Your method counts "A","K" and "K","A" separately. But that's not right; these are combinations, so they are the same. Hence what you want to do is choose two different pairs out of 13, so (13 2). Note that you could just divide your answer by 2: (13 1)(12 1)/2 = (13 2).

 

[walk_w/o_aim]

Thanks, hgfalling. I got it now. Picking one at a time then multiplying the total number of combinations with each other makes the order matter -- I should have known that from the largest number n bits can represent or something similar. Thank you.