# Generating function for picking j balls without replacement from an urn

In an urn, each balls is labeled with one of $\{0,1,2,...,k\}$. For each $i\in{0,1,2,...,k}$, there are exactly $n_i$ balls labeled $i$. Let $f(x)=\sum\limits_{i=0}^k n_ix^i$. Let $g(x)=\sum\limits_{i=0}^{jk}a_ix^i$ where $a_i$ is the number of ways drawing uniformly randomly without replacement $j$ balls from the urn with a sum of $i$.

For drawing with replacement $g(x)=f(x)^j$. Is there a nice generating function $g$ dependent on $f$ for the case of drawing without replacement?

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The first sentence is sort of unclear. The urn has $\sum_{i=0}^{k}n_i$ balls in it, not $n_i$, right? – user2357112 Jun 17 '14 at 21:38
Are two balls with the same label considered distinct? – user2357112 Jun 17 '14 at 21:42
@user2357112 Judging from the OP's statement that drawing with replacement gives $g=f^j$, it seems the answers to your questions are "yes" and "yes." – angryavian Jun 17 '14 at 21:48
@angryavian's answer is what I intended. user2357112, however, I have edited the problem statement to make it clearer. – Hans Jun 17 '14 at 22:56

Let $B$ be the set of balls and $\mathrm{label}(b)$ be the label of a ball. Then we have the following multivariable generating function:

$$\prod_{b\in B}(1+xy^{\mathrm{label}(b)})$$

The number of ways to choose $j$ balls with a sum of $i$ is the coefficient of $x^jy^i$.

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Nice. Is there an even simpler construction, say with just one variable? – Hans Jun 17 '14 at 22:54
@Hansen: Not that I could find. There doesn't seem to be a simple way to build the values for a given $j$ without using the values for lower $j$. – user2357112 Jun 18 '14 at 0:23