Probability of getting a full house

A full house has three cards of one kind and two of another, so think about it like this: first you choose a type of card (13 choices), then you choose three out of four of those cards, then you choose a second type of card, and finally you choose two of those four cards. Thus you have ${13\choose 1}{4\choose 3}{12\choose 1}{4\choose 2}$ possible full house hands. So the probability is then

$${{{13\choose 1}{4\choose 3}{12\choose 1}{4\choose 2}}\over{52\choose 5}}={{(13)(4)(12)(6)}\over2598960}={3744\over2598960}\approx0.00144$$

@Arthur Skirvin's answer is great, but I just want to provide another thinking process for the numerator:

$$(\textrm{choose two kinds})\cdot(\textrm{choose the kind of three cards}\cdot\textrm{choose cards})\cdot(\textrm{choose the kind of cards}\cdot\textrm{choose cards})\\ = {13\choose2}\cdot{2\choose1}{4\choose3}\cdot{1\choose1}{4\choose2}.$$