Bose-Einstein counting method The question I'd like to ask stems from De Mere's puzzle which is a basic problem in probability. A six-sided die is tossed three times and we are asked to judge which event is more probable : a sum of 11 or a sum of 12? The approach taken in every book I've read up to this point is to write down all the three number series of numbers that meet the requirement and divide by total number of possible outcomes. So far so easy. What would happen though if number of tosses were much higher (say 100) and we were asked to calculate $A=$ {probabilty of getting 453} ? In my mind this would boil up to counting number of solutions (tuples $(x_1,x_2,...,x_{100})$) satisfying: 
$$x_1+x_2+...+x_{100}=453$$ 
Let us also assume that order within tuples does not matter (I realise it goes against the exercise as it calls for ordered tuples but I need this kind of setting). That's kind of problem that I suppose in typical situation would be done by introducing Bose-Einstein coefficient $n+k-1\choose k$ with $n$= 453, $k$= 100. It would be wrong though as the experiment is based on rolling a die and thus we need an additional constraint: 
$$x_i \leq 6 \space {for} \space {x_i}={1,2,..,100} $$
Maybe it's based on some elementary method but I don't recall seeing it everywhere. Could someone explain how to account for existence of constraints in this method of counting?     
 A: You can attack such problems using generating functions. The idea is then to translate a counting problem with restrictions to another problem without restrictions but where you do a weighted counting and the weights keep track of certain variables that are related to the restrictions that you need to implement in the original problem.
In this case the restrictions on the variables are that they sum to 453 and that they are between 0 and 6. The latter constraints are not a big deal if the former constraint wasn't there. So, we just need to keep track of the sum of the variables while summing the variables from 0 to 6 independently of each other. This leads us to consider the function:
$$f(y) = \sum_{x_1,\ldots x_n}y^{\sum_{k=1}^{n}x_n}= \left(\frac{y^7-y}{y-1}\right)^n$$
The answer to the problem is then the coefficient of $y^{453}$ in the series expansion of $f(y)$ The question is then how to extract that coefficient. You can calculate it exactly using efficient series expansion techniques, you can also use asymptotic methods to approximate the coefficients of large powers of $y$. E.g. you can start from the contour integral representation of the coefficient of $y^n$:
$$c_n = \frac{1}{2\pi i}\oint_C\frac{f(z)}{z^{n+1}}dz$$
where the contour $C$ encircles the origin avoiding any other singularities of the integrand. In general, you can then use e.g. the saddle point method where yo deform the contour so that it moves through saddle points, allowing you to come up with a Gaussian approximation for the integral.
