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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.

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  • $\begingroup$ The $(1-x)^{-k}$ is a formal power series, and $x$ is an indeterminate so we never intend to evaluate it and thus we don't worry about convergence. $\endgroup$ – N. Shales Apr 28 '17 at 10:32
  • $\begingroup$ Sorry, exponent is $-n$ not $-k$. $\endgroup$ – N. Shales Apr 28 '17 at 10:54
  • $\begingroup$ @N.Shales, yes, we don't intend to evaluate x. This is probably the confusing / troublesome part for me. For example I had the specific problem for s = 6 and n = 3. I plucked it all in and found that, if I choose l=0 to l=15 for the 2th series, I get the right coefficients, the same as if I had expanded the series $(x + \dots x^6)^3$ . However I get infinitely more terms, terms $x^t$ with $t>18$. For me it is hard to look at these two terms and recognize that they lead to the same, correct result. Why does the 2th formal power series give me the right coefficients? $\endgroup$ – Imago Apr 28 '17 at 16:23
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    $\begingroup$ Do check out the wiki link in my first comment. There is a whole carefully defined ring structure to formal power series, so that all the results can be derived treating $x$ as indeterminate. So, for example, $(1-x)^{-1}$ is defined as the multiplicative inverse of $1-x$ it is easy to verify that $(1-x)^{-1}=1+x+x^2+x^3+\cdots$ by multiplying $(1-x)(1+x+x^2+x^3+\cdots)$ $\endgroup$ – N. Shales Apr 28 '17 at 17:43
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    $\begingroup$ @N.Shales, I begin to understand. First time I mistook the wiki page of something else and were not related too much to my problem. I have never encountered that topic before. It's completely new to me. I am srry, if I appeared unfriendly. $\endgroup$ – Imago Apr 28 '17 at 17:54
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My problem however here is that the generating function here intermixes arithmetic rules with representations as actually just the exponents and the coefficient are interesting.

That's exactly what generating functions are about: exponents represent the values of some parameter and the coefficient represents "how many cases". The "x" is just a placeholder at which to attach the exponent and the quantity. The method of generating functions works for problems that "fit" with power series arithmetic.

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