Math Forum :: View topic – Number of solutions (in N) of an equation

In relation with a probabilistic problem for concrete numbers I came upon the generalization of an idea and there’s a problem:

Let’s suppose we are throwing by special cubes with sides numbered serially by 1,2,…,m and the probability of toss of each number is the same according to the classical definition of probability). Let denote the sum of all numbers on the sides of all cubes shortly after the last toss. Which sum has the highest probability.

Let be random variables that represents the value of the toss on cubes in this order and . Then should be the value of the most probable sum on the sides of cubes (or in case that isn’t integer let’s take the closest two integers which E X is situated between). At last it’s clear that , then:

But I would like to solve it more classically and exactly, so first I need to solve this:

Let be some natural numbers such that , then the esencial question is how many (orderly) somes does there exist such that each x_i is a natural number from 1 to maximally and:

Could anybody give me any analytic formula (depending on ) for number of all convenient some (or any hint).

Obviously, the number of all convenient somes is also the number of all adic combinations with reprise of elements such that each element can appear minimally once and maximally times.

I’ve tried to use the principle of inclusion and exclusion, but it doesn’t give me the right result.

Thank you….