Suppose we want to find the number of non-negative integral solutions to the equation:
$$x_1 + x_2+ x_3 = m$$
where we have $x_i \le L_i, i\ge2$
I found the solution as:
$$\sum_{x_2=0}^{L_2} \sum_{x_3=0}^{L_3} \frac{m!}{x_2!x_3!(m-x_2-x_3)!}$$
My questions are two-fold:
Is my answer correct?
This solution is computationally impossible for $m=100$, for example. Can somebody provide the answer in a computable format?
Hint for partial checks on your solution:
If $\ \ \forall i\in\langle 1,r\rangle \ \ \ L_i = m \quad,$ then the number of non-negative solutions (that is,$\ \ 0 \le x_i \le m)$
to the equation $\quad "x_1 +x_2 + x_3 + \cdots + x_r=m" \ \ is \ \ {m+r-1\choose r-1}$
I would say if your solution does not meet this test then it is not right.