Given a number of length $l$, the sum of the digits range from $0$ to $9l$


For each sum, there are $x$ permutations of digits. find the sequence of length $9l+1$ that solves $x$.


I found a function that solves this problem without expanding the polynomial equation $l$ times:


$f(n \in S,l,b) = \sum\limits\_{i=0}^{\lfloor\frac{n}{b}\rfloor}(-1)^{i}{l \choose i}{n+l-1-bi \choose l-1}$


($l=l$, $b=10$,$S={0,1,...,9l}$)


Where $f(n,l,b)$ is the $n^{th}$ coefficient of a polynomial of length $l$ and base $b$ where $S={0,1,...,(b-1)l}$


(thanks to Tom De Medts for giving me the proper terms.)


Are there other functions that give the $n^{th}$ polynomial coefficient? I can't seem to find one.



How many permutations for each sum of digits of a number of length l? (solved but asking for insight)
Your sequences are simply the coefficients of the polynomials
$$(1 + x + x^2 + \dots + x^9)^L .$$


(This is a straightforward application of the theory of generating functions.)


