Let $n$ and $k$ be two natural numbers. In how many ways is it possible to choose $k$ natural numbers $i_{1}, i_{2}, \ldots, i_{k}$ such that $i_{1}+i_{2}+\ldots+i_{k}=n$?
The idea is to slightly transform the problem. We consider a series of $n+(k-1)=n+k-1$ cells in which we will place $k-1$ delimiter cubes. Each of these configurations corresponds to a sum: $i_{1}$ is the number of cells before the first delimiter cube and the number $i_{k}$ is the number of cells between the $(k-1)^{\text{th}}$ delimiter and the end of the row. Conversely, for each sum, there is a unique arrangement of cells and delimiter cubes.

![](https://cdn.mathpix.com/cropped/2024_05_10_60ac669be8d11ef61a82g-63.jpg?height=152&width=998&top_left_y=1884&top_left_x=214)

FIGURE 1 - An example for $n=7$ and $k=5$, the corresponding sum is $7=0+3+0+3+1$

Thus, there are as many sums of $k$ numbers whose sum is $n$ as there are choices of $k-1$ cells out of $n+k-1$. Therefore, there are $\binom{n+k-1}{k-1}$ possible sums.