# Why am I under-counting when calculating the probability of a full house?

I was trying to answer this question.

Find the probability of getting a full house from a $52$ card deck. That is, find the probability of picking a pair of cards with the same rank (face value), and a triple with equal rank (different from the rank of the pair of course).

Take one rank. The number of ways you can get a pair from a rank is $C(4, 2)$.

Take another rank. The number of ways you can get a triple from that rank is $C(4, 3)$.

Therefore, the number of ways you can get a full house from just $2$ ranks is $C(4, 2) \cdot C(4, 3)$. But there are $13$ different ranks, and the number of ways you can select $2$ ranks out of $13$ is $C(13, 2)$. For each of these pairs of ranks, you can have $C(4, 2) \cdot C(4, 3)$ full houses. Therefore the number of possible full houses is $C(4, 2) \cdot C(4, 3) \cdot C(13, 2)$, and the probability of selecting a full house is

$\dfrac {C(4, 2) \cdot C(4, 3) \cdot C(13, 2)}{C(52, 5)} \approx 0.000720288$

However, this is way lower than the actual answer of $.00144$. Where did I go wrong here?

You're missing a factor of $2$ because you're not distinguishing between the two ranks. It makes a difference whether you have $2$ of one and $3$ of the other or the other way around.