What is the probability of getting two pair (ex: two 3's, two 7's and a ace) from a random 5 card poker hand?

Question

I know this question has been asked several times. But I appraoched the problem in a different way and it leads me to a different answer. My apprach is as follows: $$ \frac{\binom{13}{3}\binom{4}{2}\binom{4}{2}\binom{4}{1}}{\binom{52}{5}}\ $$ That is, first I am choosing three different denominations out of 13 and then for the first 2 cards I will be selecting 2 suits out of 4 and similarly for 3rd and 4th card. And for the last card I can choose any of the 4 suits. But the answer I get is one-third of the original answer. original answer $$ \frac{\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1}}{\binom{52}{5}}\ $$ What is the flaw in my analysis? Thank you

Answer 1

Using $\binom{13}3$ means you are choosing three denominations with order not important. So you also need to choose which of your three denominations is the single card. There are $3$ possibilities, so your answer needs to be multiplied by $3$.