From a standard deck (52 cards, 4 suits) of playing cards, how many ways is it possible to be dealt three kings as a three of a kind? So the two other cards cannot be able to form a pair nor can they be another king as that would make it a four of a kind.
This is the answer I came up with, however, I feel like it is wrong. There is a possibility that this expression allows me to pick any three of a kind.
$$[C(13,1)\cdot C(4,3)] \cdot [C(12,2)\cdot C(4,1) \cdot C(4,1)]$$
To prevent such occurrences, would I have to divide the expression with "13!"? Or would I have to rewrite the expression in a completely new way? I might be overthinking this but I still feel like I am wrong.
First handle the king:$\binom{4}3$
and then choose $2$ cards out of the other denomination:$\binom{12}2$.
Choose the suit for the two cards: $4^2$.
Hence overall, $$\binom{4}3\binom{12}24^2=4^3\cdot \frac{12(11)}{2!}=(66)(64)=4224$$
Remark: There is no need to have $\binom{13}1$ since we already know we want that to be the king.