# There's a deck of 52 cards. A five-card hand has three of a kind consists of 3 cards of the same rank, one card of another, and one card of another. How many different 3-of-a-kind hands are there?

$\left(\begin{matrix}13 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 3\end{matrix}\right) \left(\begin{matrix}12 \\ 2\end{matrix}\right) {\left(\begin{matrix}4 \\ 1\end{matrix}\right)}^{2} = 54 , 912$

#### Explanation:

To get a three-of-a-kind, we have 3 of one ordinal, and 2 other cards that are not the same ordinal as the first three or each other.

In all of the calculation, we're using combinations (we don't care about the order of the card draw):

C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

There are 13 ordinals in a deck of cards (Ace, or 1, through 10, and then Jack, Queen, and King). We are picking one of the ordinals:

$\left(\begin{matrix}13 \\ 1\end{matrix}\right)$

and from that ordinal we want 3 of the suits (there are 4 suits for each ordinal - Spades, Hearts, Clubs, Diamonds):

$\left(\begin{matrix}13 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 3\end{matrix}\right)$

We then need to get 2 more cards. They can't be of the same ordinal as the first one or each other (so that's 12 remaining ordinals and we're picking 2). We want one of each:

$\left(\begin{matrix}13 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 3\end{matrix}\right) \left(\begin{matrix}12 \\ 2\end{matrix}\right) {\left(\begin{matrix}4 \\ 1\end{matrix}\right)}^{2}$

and now we can calculate this:

$13 \times 4 \times 66 \times {4}^{2} = 54 , 912$