If it is assumed that all ${52 \choose 5}$ poker hands are equally likely, what is the probability of two pairs? (This occurs when the cards have denominations $a,a,b,b,c$, where $a,b,c$ are all distinct.
So my logic was
$$\frac{{13\choose 1}{4\choose 2}{12 \choose 1}{4\choose 2}{11\choose 1}{4\choose 1}}{{52\choose 5}}$$
but the answer is:
$$\frac{{13\choose 2}{4\choose 2}{4 \choose 2}{11\choose 1}{4\choose 1}}{{52\choose 5}}$$
My logic was that for the first pair you need to choose the value of the pair from 13, then choose the 2 suits for those cards. then you have 12 values left for the second pair to choose from, and again choose 2 suits for those cards.
But in the solution it does ${13\choose 2}$ which makes sense. but im just not understanding why both solutions aren't right. it makes sense that you choose 2 values for the two pairs from 13 total values. Is my answer wrong because I am assuming order matters? i.e. I am first choosing one pair and then another?
