Number of possible solutions to equation

Question

I am trying to solve $$x+y+z = 32$$ Where $x$, $y$, and $z$ are positive integers

I believe the answer is: $C_{2}^{31}=465$ but I am not sure why. Can someone please explain?

Answer 1

Writing $32$ as the sum of three positive integers is equivalent to writing $32 - 3 = 29$ as the sum of three non-negative integers, so let's count the solutions of $29 = x'+y'+z'$ where $x',y',z'$ are non-negative integers. The solutions of this equation correspond to the number of possibilities you have when picking $29$ times one of three different objects, where repetitions are allowed and order doesn't matter. By the "well known" stars-and-bars argument (see here), the number of these possibilities is indeed ${29+3-1\choose3-1} = {31\choose2},$ as claimed.