Explicit form for sum of "multichoose" functions.

Question

Suppose I have to draw $k$ balls from a set of $n$ balls with replacement, but without order. It is known that the number of ways to do that is given by $$ \binom{n+k-1}{k}. $$ Question: Does there exist an explicit form for the number of combinations to draw $\ell$ or less balls, i.e. $$ \sum_{k=1}^{\ell}\binom{n+k-1}{k}. $$

Answer 1

Note that $$ \sum_{k=1}^{\ell}\binom{n+k-1}{k}=\sum_{k=1}^{\ell}\binom{n+k-1}{n-1}=\sum_{i=n}^{n+\ell-1} \binom{i}{n-1}=\sum_{i=n-1}^{n+\ell-1}\binom{i}{n-1}-1 $$

Using the Hockey Stick Identity, we know that $$\sum_{i=n-1}^{n+\ell-1}\binom{i}{n-1} =\binom{n+\ell}{n}$$ So the answer is $$\binom{n+\ell}{n}-1$$

Answer 2

Choosing $l$ or fewer balls from $n$ is the same as choosing exactly $l$ balls from $n+1$, then throwing away any copies of the last ball, except that this includes the case of having $0$ balls left, and you don't want to count that. So your answer is (number of ways to choose $l$ from $n+1$)$-1$.