Suppose we have a list of digits, which has $n_1$ digits of kind $a$, $n_2$ digits of kind $b$, $n_3$ digits of kind $c$. How can I get the number of numbers with $n_1 + n_2 + n_3$ digits that can be made out of the digits from the list?
Each digit from the list can be used just once, so each number will have only $n_1$ digits of kind a; however, there is no difference between all the digits of the a kind, all the digits from the $b$ kind and so on. So $ab_1b_2c$ is just the same as $ab_2b_1c$.
(Example: List: 1, 2, 2, 3; Possibilities: 1223, 1232, 1323, 1332, 2123, 2132, 2213, 2231, 2313....)
The idea is to consider at first the overall number of permutations that is
$$(n_1+n_2+n_3)!$$
and then recognize that for each kind we have $n_i!$ identical permutations therefore finally we obtain
$$\frac{(n_1+n_2+n_3)!}{n_1!n_2!n_3!}$$