# $x_1 + x_2 + x_3 +\cdots + x_m = k$

What I'm tyring to show is the number of solutions to the equation of natural numbers;

$$x_1 + x_2 + x_3 +\cdots + x_m = k$$ is equal to $$\binom{m + k - 1} m$$

To be blunt, I have no idea where to start

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Consider a string with $k$ symbols, which you need to partitition into $m$ regions, say by inserting $m-1$ separator characters "|". That gives your total string of length $k+m-1$, of which you need to pick where the $m-1$ separators go.
the total number of ways to do that is $k+m-1\choose{m-1}$