A random 5 card poker hand is picked from a standard deck of cards. Find the probability of getting two different pairs(e.g., two 3’s, two 7’s, and an Ace).
The answer I found online: $$\frac{{\binom{13}{2}.\binom{4}{2}}^2.44}{\binom{52}{5}}$$
But, I think it should be: $$\frac{{\binom{13}{2}.\binom{3}{1}}^2.44}{\binom{52}{5}},$$ where $$\binom{13}{2}\text{ is the number of ways of selecting any 2 cards from a suit, i.e., 13 cards, and}$$ $$\binom{3}{1}\text{ is the number of ways of selecting 1 card out of remaining 3 same card to form a pair}$$ $$\binom{3}{1}^2\text{denotes ways of selecting both pairs, and}$$ $$\text{44 - total number of ways of selecting 5th card which doesn't belong to any pair, and}$$ $$\binom{52}{2}\text{ is the number of ways of selecting 5 cards out of total 52 cards}$$
Can someone please help me figuring out the correct one? Thank in advance.
The event is to obtain two from thirteen kinds each of two from four suits$^\star$, and one from eleven kind of one from fourty-four cards not of those kinds; when selecting five from all fifty-two cards.$$\dfrac{\dbinom{13}2\dbinom 42^2~\dbinom {11}1\dbinom41}{\dbinom{52}{5}}$$
$\star$ You must obtain the two of each kind and both suits for each of them.