Number of outcomes, if having known the distinct numbers and number of choices This question came to me, when I was solving another relavent question in my class:
We have $N$ distinct numbers, say $P(X=i)=1/N$, with $i=1,...,N$. We choose $n$ (known) numbers from them (with replacement). But we do not know what we have got excatly for each time. What we only know is the distinct numbers in our choices $(X_{1}^{*},...,X_{r}^{*})$ (Update: No order in them, which means $X_{i}^{*}$ could happen at any time within $n$ tries), and the size of this set $r$. The question is to calculate the number of outcomes given known $(X_{1}^{*},...,X_{r}^{*})$ and $r$. Or the following probability:
$$P(X_{1}^{*},...,X_{r}^{*},r,n|N)$$
 A: So we know that we sampled $n$ times, and we know the numbers that came up, but not their respective multiplicities. 
Arrange the numbers we got in order, and let their multiplicities be $x_1$ to $x_r$ respectively. Then $x_1+\cdots +x_r=n$. By Stars and Bars the number of possibilities is $\binom{n+r-1}{n}$.
Note that these possibilities are not all equally likely.
Added: For order matters, we are counting the number of onto functions from the set $\{1,2,\dots,n\}$ to the set $\{1,2,\dots,r\}$. For information on how to count this, please see Stirling Numbers of the Second Kind. There is not a closed form, but there are useful recurrences.
A: So you want to know the probability of obtaining $r$ specific distinct numbers by unbiased selection of $n$ from $N$ numbers with replacement.
The Stirling number of the second kind $\begin{Bmatrix}n\\r\end{Bmatrix}$ counts the ways to partition $n$ labelled items into $r$ non-empty unlabelled subsets.
$$P=\dfrac{\begin{Bmatrix}n\\r\end{Bmatrix}r!}{n^N}$$
