There are a number of ways to describe this problem. I shall name a few.
Submultisets
Let $(M, f)$ be a multiset where $M = {x_1, … x_k}, |M| = k$ and $f(x_i) = m_i$, i ranging from 0 to k inclusive. Find the number of submultisets $(S,g)$ of $(M,f)$ of size n, $|(S, g)| = n$.
Integer Partitions
Let $n, x_1, … x_k, m_1, … m_k \in \mathbb{N}$
Find the number of solutions of the equation:
$x_1 + x_2 + … x_k = n$
such that $0 \le x_i \le m_i$, i of course ranging from 0 to k inclusive.
Unordered combinations of size n
Let a collection of objects M have k types of objects, where there are $m_i$ identical objects for each type, i from 0 to k inclusive. How many ways can I pick n elements from M, given the restriction that I cannot pick more than $m_i$ from each type?
Unless I've made a mistake somewhere, each of these problems should be identical.
In the ordinary undergrad combinatorics literature, there is a method for picking the number of ways to solve this, without the $m_i$ restrictions I mentioned (or at least $m_i \ge n$. That is call the multiset coefficient. There is also the number of ways of ordering all of the elements of M, provided you use all of them. That is called the multinomial coefficient. This, of course, uses only n elements from M, and does not include ordering.
The only solution I've come across, given the 5 measly hours I've researched the problem, was in Combinatorics and Graph Theory by Harris et. al, which mentions a crude method for counting the number of strings of length 4 of a given set of letters "Alma, Alabama". They enumerate 4 different patterns by hand, and use well known equations on those patterns. It works when n and k are small, but is inefficient otherwise.
Update
Sorry, I should have made my question a bit more explicit. I'm looking for an efficient method to do it by hand. The recurrence equations are helpful, but it's not quite what I was looking for.
I don't think there is a simple equation that provides the answers. A recurrence equation is likely to catch all of the cases, but may not be efficient for computing by hand. I don't think there is an efficient way to do this, if k is large and n is small. Or is there?
Update 2
Forgive me for bringing up the concept of a multiset. I thought it was in the standard literature. The size of a multiset $|(M, f)|$ is just the summation of the multiplicities $|(M, f)| = \sum_{i} f(x_i)$. It's not the same as the size of the support $|M|$. They are only equal when $f(x_i) = 1$ for all of the elements, just a regular set.
Best Answer
I don't really understand the description of the SubMultiset problem.
I think I do understand the description of the integer partition problem.
This can be solved by generating functions.
The number of solutions to
$x_1 + x_2 + \dots + x_k = n$ with $0 \le x_i \le m_i$ is the coefficient of $x^n$ in
$$ (1+x+\dots+x^{m_1})(1+x+\dots+x^{m_2})\dots(1+x+\dots+x^{m_k}) $$
$$ = \frac{\prod_{i}^{k}(1-x^{m_i+1})}{(1-x)^{k}}$$
$$ = (\sum_{j=0}^{\infty}{k+j-1 \choose j} x^j) \prod_{i}^{k}(1-x^{m_i+1})$$