Given the problem from: extended stars-and-bars problem(where the upper limit of the variable is bounded)

If $\mathbf{r} = (r_1, r_2, \ldots, r_n)\in \mathbb{N}^n_0$, i.e. $r_i$ is a non-negative integer for each $i=1,\ldots,n$, what is the number of solutions $\mathbf{a}=(a_1,a_2,\ldots,a_n) \in \mathbb{N}^n_0$ satisfying $|\mathbf{a}| = \sum_{i=1}^n a_i = N$ and $a_i \leq r_i$ for each $i=1,\ldots,n$.

I would like to find a sharp bound for the number of solutions to this problem which holds for each $\mathbf{r}\in \mathbb{N}_0^n$ such that $|\mathbf{r}|=j$ for some fixed $j\in \mathbb{N}_0$. The bound should also be dependent on $N$. Let $C(\mathbf{r},N)$ denote the coefficient in the polynomial function \begin{align*} P(X) = \frac{\prod_{i=1}^n (1-X^{r_i+1})}{(1-X)^n} \end{align*} for $X^N$. Is it true that \begin{align*} C(\mathbf{r},N) \leq { j \choose N } \end{align*} for every $\mathbf{r}\in\mathbb{N}_0^n$ such that $|\mathbf{r}|=j$? Is there a sharper bound? Note that when $\mathbf{r} = (1,1,\ldots,1) \in \mathbb{N}^n_0$ and $|\mathbf{r}| = n$, then ${ |\mathbf{r}| \choose N } = C(\mathbf{r},N)$.

Proof: When $\mathbf{r} = (1,1,\ldots,1)$ and $|\mathbf{r}|=n$, then \begin{align*} P(X) = \frac{\prod_{i=1}^n (1-X^2)}{(1-X)^n} = \frac{((1+X)(1-X))^n}{(1-X)^n} = (1+X)^n = \sum_{k=0}^n { n \choose k } x^k \end{align*} by the binomial theorem, and since $n = |\mathbf{r}|$, we see the result is true for the $N$-th term in the sum.

I would also like to obtain a reference or derivation for the polynomial function \begin{align*} P(X) = \frac{\prod_{i=1}^n (1-X^{r_i+1})}{(1-X)^n} \end{align*} given by user Marc van Leeuwen in counting solutions to this problem.


Yes $C(r,N) \leq { |r| \choose N }$ as it is easy to come up with a surjective function from the selection ${ |r| \choose N }$ of balls to solutions $a_1 + a_2 + \ldots + a_n = N$. Out of the $|r|$ balls, label $r_i$ of them the number $i$. Given the $N$ chosen balls, assign to $a_i$ the number of balls chosen with the label $i$. It is clear that this always leads to a solution and that every solution is attainable in atleast one way, hence $C(r,N) \leq { |r| \choose N }$.

The formula for $P(x)$ is just a simplified expression of

$$ \left(1+x+\ldots+x^{r_1}\right)\, \left(1 + x + \ldots + x^{r_2}\right) \ldots \left(1 + x + \ldots + x^{r_n}\right)$$

for which the coefficients are in bijection with solutions $\boldsymbol{a}$ by taking $a_i$ to be the exponent taken from the $i$th bracket.

A little trick that I find particularly useful when solving these types of problems. Suppose $|r| = N+\eta$ for some small $\eta$. Then by making the change of variables $r_i - b_i = a_i$ for each $i$, we find that we only need to count the number of nonnegative integer solutions $b_1 + \ldots + b_n = \eta$ where vector $r$ remains unchanged. This can be realized combinatorially as well (balls that you don't choose).

  • $\begingroup$ Could you please provide a reference for the formula for P(x)? $\endgroup$ – user129334 Jun 1 '15 at 16:01
  • $\begingroup$ @user129334 Any combinatorics book will have it as an example in their generating functions section. Perhaps an example will help clarify. Suppose N=6 and r=(3,3,3). We have P(x)=(1+...+x^3) (1+...+x^3) (1+...+x^3). Note this can be rewritten as ((1-x^4)/(1-x))^3 but ignore this form as it is less intuitive. When you multiply this out, we are looking for the coefficient of x^6. Each solution contributes to this coefficient. For example, a=(2,3,1) picks out (x^2)(x^3)(x). $\endgroup$ – Andrew Szymczak Jun 1 '15 at 17:38
  • $\begingroup$ Thank you, giving P(x) a name and your example helped me find more information. $\endgroup$ – user129334 Jun 1 '15 at 18:06

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