I had first wondered about a special case of, how to calculate the number of possibilities so that the sum of eyes of $n$ thrown dices equals a certain value $c$.
After some time thinking, I decided to google and found on mathworld the following equations describing the general case of my initial problem. For simplification I have copied the interesting lines and posted them here:
\begin{align} f(x) &= (x+x^1 + \dots + x^s )^n \tag 1 \\ & = x^n {\left( \sum\limits_{i=0}^{s-1}x^i\right)^n} \tag 2 \\ & = x^n \left(\frac{1-x^s}{1-x}\right)^n \tag 3 \\ & = x^n (1-x^s)^n \left(\frac{1}{1-x}\right)^n \tag 4 \\ & = x^n \sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} x^{sk} \sum\limits_{l = 0}^{\infty} \binom{n+l-1}{l} x^l \tag 5 \end{align}
The steps down from $1-4$ are not troublesome at all. From step $4$ to $5$ one uses the binominal theorem and then its generalized version.
Using the gerenalized version of the binominal theorem however requires the condition: $x < |1| $
It had not been stated on mathworld, but obviously assumed to do that step. T My question now: How do we assume that and why? What do the $x$ mean in this generating function, if one has to add additional condition on them?
My guess:
Obviously the exponents in $x, \dots, x^s$ are simply the representation of the numbers $1$ to $s$ on our dice and that if the dice is fear, then the whole expression $x, \dots, x^s$ represent the resprective likelihoods that a certain result appears in a throw and then making their coefficients of the number of possible combinations to get to that result, well, of course the likelihoods are $\in [0,1]$ . My problem however here is that the generating function here interemixes arithmetic rules with representations as actually just the exponents and the coefficient are interesting.