Probability of a five-card poker hand containing 2 pairs I am trying to solve this question in my own way. I tried counting like this-
First card can be chosen in 52 ways, second in 3 ways because it has to be of same kind as first, third in 48 ways, fourth in 3 ways and fifth in 44 ways.
I got the answer as :
(52*3*48*3*44)/5! as the favorable events.
I was trying to solve it like the way it is solved here
Can anyone help me get it correct because I am not getting the correct answer this way

If you start with the pairs, you're going to run into trouble since at some point you have to divide by multiple $2$s to account for swapping cards within pairs, as well as swapping the pairs. Rather, you should start with the kicker, and also fix one property at a time and not one card at a time:

*

*$13$ ways to select the rank of the kicker, $4$ choices for its suit

*$\frac{12×11}2$ ways to select the ranks of the two pairs

*$\frac{4×3}2×\frac{4×3}2$ ways to select the suits of each pair

