# Distributing M identical objects in N containers with capacity C

What is the number of integer solutions to $x_1+x_2+ \cdots + x_N = M$ where $0 \leq x_i \leq C$ for $i = 1, \dots, N$? (All constants are positive integers.)

• Is the order important or not? – Marc Bogaerts Sep 22 '14 at 10:07
• The order is not important; all objects are identical, all containers are similar (max C capacity). The end arrangement of the objects is important; however, the containers are indexed and can be distinguished. – Alfred Hess Sep 22 '14 at 10:11
• i.e., the variables x. – Alfred Hess Sep 22 '14 at 10:25
• So $x_1+x_2+\ldots$ is different from $x_2+x_1+\ldots$? – Marc Bogaerts Sep 22 '14 at 11:26
• No, they are the same. What I meant was that x1 is different from x2, and ... – Alfred Hess Sep 22 '14 at 12:05

## 1 Answer

If you're familiar with generating functions, you can get a pretty explicit answer. Here's a stab at it. Each container can be represented with the polynomial $1+x+\dots+x^C=\frac{1-x^{C+1}}{1-x}$, which is a sum of powers of $x$, one for each allowable number of objects in a container. Then, the way to represent the collection of all $n$ containers is by taking the product of these polynomials, namely $\left(\frac{1-x^{C+1}}{1-x}\right)^N$. Now the coefficient of $x^M$ in this product gives the number of ways to arrange exactly $M$ objects among the $N$ containers. In other words, the number of solutions to your equation is $$[x^M]\left(\frac{1-x^{C+1}}{1-x}\right)^N.$$ To extract this coefficient takes a bit of maneuvering. However, with the binomial theorem and some knowledge of the Taylor series of $\frac1{(1-x)^N}$, you should be able to do this and get the exact count you're looking for, as a sum of $N$ terms of products of binomial coefficients.