calculus - How many n-th Order Partial Derivatives Exist for a Function of k Variables?

Example:
Let's say for example I have a function, $f$, of $2$ variables :

$f(x,y)$

For this function there exits $2$ first-order partial derivatives namely:

$f_x = \frac{\partial f}{\partial x}$

$f_y = \frac{\partial f}{\partial y}$

Then if we are to differentiate further we will find that there are $4$ computable second-order partial derivatives.

$f_{xx} = \frac{\partial^2 f}{\partial x^2}$

$f_{xy} = \frac{\partial^2 f}{{\partial x} {\partial y}}$

$f_{yy} = \frac{\partial^2 f}{\partial y^2}$

$f_{yx} = \frac{\partial^2 f}{{\partial y} {\partial x}}$

However due to the Equality of Mixed Partials (https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives), two of those second-order partial derivatives are equivalent, $f_{xy} = f_{yx}$, and thus we are left with $3$ second-order partial derivatives for a function of 2 variables.

$f_{xx} = \frac{\partial^2 f}{\partial x^2}$

$f_{xy} = \frac{\partial^2 f}{{\partial x} {\partial y}} \Leftrightarrow f_{yx} = \frac{\partial^2 f}{{\partial y} {\partial x}}$

$f_{yy} = \frac{\partial^2 f}{\partial y^2}$

Question: Given a function of k variables:

$f(x_1 , x_2,x_3,\dots,x_{k-1},x_k)$

Is there a formula to find the number of $n^{th}$-order partial derivatives, (where $n$ is the order of the partial derivative), for a function of $k$ variables?

For example where $n=1$ (i.e. the first-order derivatives), there would be $k$ partial derivatives, just as in the example above, for a function of $2$ variables there exists $2$ first-order derivatives.

Answer

This is the problem of distributing $n$ balls over $k$ bins, which can be solved using the stars and bars approach; the result is

$$\binom{n+k-1}{k-1}\;.$$