Find the probability that you have three of a kind after you pick 5 cards from a regular 52-card deck. How would I go about finding this? I was thinking that, since this is a permutation, you would  have
52!/(52-5!)? But how would that account for 3 of a kind?
 A: There are $\binom{52}{5}$ five-card hands. They are all equally likely. 
We now count the favourables, the three of a kind hands. The kind we have $3$ of can be chosen in $\binom{13}{1}$ ways. For each of these choices, there are $\binom{4}{3}$ ways to choose the actual $3$ cards. 
Now we must count the number of ways to choose the "useless" cards. The two different kinds that we have one each of can be chosen in $\binom{12}{2}$ ways. For each of these ways, the actual cards of the chosen kinds can be chosen in $\binom{4}{1}\binom{4}{1}$ ways. Thus the number of favourables is $\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1}$.
For the probability, divide by $\binom{52}{5}$.
Remark: The following is an alternate approach to count the number of ways to choose the useless cards. There are $\binom{48}{2}$ ways to choose $2$ cards from the $48$ cards that are not of the kind we have $3$ of. But some of these choices give us $2$ of the same kind, giving us a full house, a very good hand. So we must subtract the number of ways in which we get $2$ of a kind. The kind can be chosen in $\binom{12}{1}$ ways, and the actual cards in $\binom{4}{2}$ ways. So the number of ways to choose the useless cards is $\binom{48}{2}-\binom{12}{1}\binom{4}{2}$. 
