What is the number of $m$-tuples $\left(x_{1}, x_{2}, \ldots, x_{m}\right) \in\left(\mathbb{N}^{*}\right)^{m}$ satisfying $x_{1}+\ldots+x_{m}=n ?$
For each $m$-tuple $\left(x_{1}, \ldots, x_{m}\right)$ of strictly positive integers such that $x_{1}+\cdots+x_{m}=n$, we can associate the following representation.

![](https://cdn.mathpix.com/cropped/2024_05_10_8c44cd8b5c585c15e58bg-431.jpg?height=240&width=1105&top_left_y=914&top_left_x=316)

Choosing a suitable $m$-tuple is equivalent to choosing the positions of the $m-1$ separators. There are $n-1$ possible positions. The answer is therefore $\binom{n-1}{m-1}$.