There are $\binom{52}{5}$ ways to choose $5$ cards from $52$. All these ways are equally likely. Now we will count the number of "full house" hands.
For a full house, there are $\binom{13}{1}$ ways to choose the kind we have three of. For each of these ways, the actual cards can be chosen in $\binom{4}{3}$ ways. For each way of getting so far, there are $\binom{12}{1}$ ways to choose the kind we have two of, and for each there are $\binom{4}{2}$ ways to choose the actual cards. So our probability is $\frac{\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}}{\binom{52}{5}}.$
Remark: We have used binomial coefficients systematically, even when a simpler expression was available. For example, there are clearly $13$ ways to choose the kind we have three of.
To calculate the binomial coefficient $\binom{n}{k}$, a reasonably efficient procedure, a not too bad way, when $k$ is not large, is to use $\binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}.$ For $\binom{52}{5}$ the numerator is $(52)(51)(50)(49)(48)$ and the denominator is $(5)(4)(3)(2)(1)$.
A similar procedure can be used to find the probabilities of the other standard poker hands. The only place where a mistake is fairly common is the probability of two pairs.
For example, to count the number of one pair hands, do this. The kind we have a pair of can be chosen in $\binom{13}{1}$ ways, and for each of these ways the actual cards can be chosen in $\binom{3}{2}$ ways. Now the three kinds we have one each of can be chosen in $\binom{12}{3}$ ways, and the actual cards can be chosen in $\binom{4}{1}\binom{4}{1}\binom{4}{1}$ ways, for a total of $\binom{13}{1}\binom{4}{2}\binom{12}{3} \binom{4}{1}\binom{4}{1}\binom{4}{1}$.