How many different numbers are composed by n repeated digits? For example, there are 3 digits: 1, 1, 4 and they compose 3 different numbers: 114, 141, 411.
My questions is: given n repeated digits: 1 * n1, 2 * n2, 3 * n3, ..., 9 * n9, in which ni >= 0 and n1 + n2 + ... + n9 = n, how many different numbers are composed by the n digits?
 A: Usually, the easiest way to answer these questions is to think as follows:
1) Suppose that we can tell ALL of the objects apart.  (For instance, say we numbers the 1's as $1_1,1_2,\ldots,1_{n_1}$, etc.).  How many ways could these be arranged?  In this case, that is just $n!$, of course.
2) Obviously this was an overcounting.  But by how much? In other words, how many rearrangements of the "labeled" objects should all count as the same "unlabeled" object?  Given a sequence, you can rearrange the 1's in any way you like, as long as the positions that have 1's don't change; since there are $n_1$ 1's, this can be done in $n_1!$ ways.  Similarly for the rest.
So, overall, there are
$$
\frac{n!}{n_1!n_2!\cdots n_9!}=\binom{n}{n_1,\ldots,n_9},
$$
where this last is the multinomial coefficient.
A: I take it you are asking about numbers that use all the available digits. If you are familiar with Multinomial Coefficients, the answer is immediate: By definition the number of choices is
$$\binom{n}{n_1,n_2,\dots, n_9}.$$
This is equal to
$$\frac{n!}{n_1!n_2!n_3!\cdots n_9!}.$$
A: First you pick the positions for the $n_1$ digits in ${n \choose n_1}$ ways, then you pick the positions fo the $n_2$ digits in ${n-n_1 \choose n_2}$ ways and so on.  Multiply them and you are there.
