# Calculating probability that you are dealt a full house [duplicate]

This question already has an answer here:

What is the probability that you are dealt a "full house"? (Three cards of one rank and two cards of another rank.)

why can't I calculate the probability in the following way?

The number of ways I can select 2 suits from one rank = $$13 \choose 2$$ The number of ways I can select 2 suits from one rank = $$4 \choose 2$$ The number of ways I can select 3 suits from another rank = $$4 \choose 3$$

so p should be $$\frac{{13 \choose 2}{4 \choose 2}{4 \choose 3}}{52\choose5}$$

My question is how can I tweak my approach and reach the right answer.

## marked as duplicate by NCh, Zain Patel, Daniel W. Farlow, Shailesh, The Dead LegendMay 6 '17 at 2:40

$\binom{13}{2}$ is not the right choice to make, because the two ranks you pick do not have the same role in the rest of the computations, so for example "sevens and queens" is a different choice at this stage than "queens and sevens".
Thus, instead of $\binom{13}{2}$ you should just use $13\cdot 12$ -- namely, one factor for the rank there'll be three of, then one factor for the rank there'll be two of.