Number of distinct derivatives of order $n$ of an $m$-variable function

Question

Advanced Calculus 2nd ed., by David Widder, says that the number of distinct (partial) derivatives of order $n$ of a (presumably smooth) function of $m$ variables is given by

$$\binom{n+m-1}{n} = \frac{(n+m-1)!}{n! (m-1)!}.$$

I drew out a tree diagram with the root being the original function $f$, and whose children as $f_{1}, \cdots, f_{m}$ representing the first partial derivatives of $f$ wrt to each input. Each of these, $f_i$ have children as well, given by $f_{i,1}, \cdots, f_{i,m}$. And these have children as well, etc.

This tree-like diagram gave me an enjoyable visual that suggests for a given order we must count the number of leaves of an $m$-furcating tree of depth $n$ from the root.

I have tried to think about the addition and product rules of counting with little success. How do I derive this counting equation?

Answer 1

Let $y_i$ be the number of times the derivative is taken with respect to $x_i$. You want to count the nonnegative integer solutions to $\sum_{i=1}^m y_i = n$. By stars and bars, the count is $\binom{n+m-1}{n}$.