Three-of-a-kind Poker Hand Problem Problem:

*

*Three-of-a-kind poker hand: Three cards have one rank and the remaining two cards have
two other ranks. e.g. {2♥, 2♠, 2♣, 5♣, K♦}

Calculate the probability of drawing this kind of poker hand.
My confusion: When choosing the three ranks, the explanation used $13 \choose 1$ and $12 \choose 2$. I used $13 \choose 3$ instead which ends up being wrong. I do not know why.

Here is another way to solve it through unordered samples.
We are looking for hands of the kind $x_1$-$x_2$-$x_2$-$y$-$z$, where $x_1,x_2,x_3$ are all of the same face value (although of a different suit), whereas $y,z$ are different face values.
To work with unordered hands, let's fix the order of the cards as above, i.e. three of a kind are the first three cards followed by two other different kinds.
There are 13 possible face values (2, 3, $\ldots$, K, A), and for each face value, there are ${4\choose 3}$ ways to select 3 cards out of 4, disregarding order and without replacement. This fills $x_1$-$x_2$-$x_3$.
For $y$, there are 48 possibilities, since three face values have already been drawn and the one left cannot be used. For $z$, there are 44 possibilities since the other three remaining cards of the face values chosen in $y$ cannot be either.
However, we are not done yet, i.e. $13{4\choose 3}48\cdot44$ is not quite right because this number includes also poker hands s.t. 4s-4c-4h-2s-3h and 4s-4c-4h-3h-2s, which are obviously indistinguishable since order doesn't matter. But the last two cards can be ordered in 2! ways. Dividing by $2!$, we remove those hands that differ only in the ordering in the last two cards.
The right number of poker hands with a three-of-a-kind is thus
$$13{4\choose 3}\frac{48\cdot44}{2!},$$
and the required probability is
$$
\frac{13 {4\choose 3}\frac{48\cdot44}{2!}}{{52 \choose 5}}.
$$
