Probability of full house

I'm trying to calcualte the probability of a full house in a standard 52-card deck.

We choose out of the thirteen kinds $2$ of them so $\binom{13}{2}$. There are $4$ houses of each card in the deck so we must choose $3$ of them for the first kind, and $2$ of them for the second kind ($\binom{4}{3}\cdot\binom{4}{2})$. Now permute this $5$ cards into al possible arrangements $5!$.

Sample space: $|S|=P(52,5)$

Event: $E=\binom{13}{2}\cdot\binom{4}{3}\cdot\binom{4}{2}\cdot5!$

This gives probability: $\frac{3}{4165}$ and I seem to be undercounting by a factor of $2$. What I'm I doing wrong?

The kind we have $3$ of can be chosen in $13$ ways. For each of these the kind we have $2$ of can be chosen in $12$ ways. Your $\binom{13}{2}$ neglected the fact that $3$ Aces and $2$ Kings is different from $3$ Kings and $2$ Aces. Or to put it another way, after choosing the two suits, in $\binom{13}{2}$ ways, we must decide which of the chosen suits will be represented by three cards, and which by two. But instead of analyzing the number of ways as $\binom{13}{2}\cdot 2$, it is easier to count them as $(13)(12)$.
Remarks: $1.$ The $\binom{13}{2}$ would be appropriate as a step in counting two pair hands. The usual mistake is not to undercount full house hands, it is to overcount two pair hands, by using $(13)(12)$ instead of $\binom{13}{2}$.
$2.$ I would probably argue that there are $\binom{52}{5}$ equally likely hands. Maybe you call it $C(52,5)$. Then in counting the favourables, we are counting hands, so we do not have to multiply by $5!$.