Permutations of $3$ elements which sum to $100$

Question

I need to find the number of permutations of $3$ elements which add up to $100$. For each element, $0 \leq x \leq 100$. So we could have $0,0,100$; $1,0,99$; $0,1,99$ etc. Order is important - I need to count $1,0,99$ as well as $0,1,99$. Can anyone help?

Answer 1

Hint:

This problem is the same as distributing 100 balls into 3 distinct boxes i.e.

$$x+y+z=100$$

Each ball (=the number $1$) is identical and the boxes ($x,y,z,$) are distinct

Answer 2

Using another formulation, you need to solve $x+y+z=100$ with $x,y,z\in\mathbb{N}$.

A hint: count all $(x,y) \in [0,100]$ such that $x+y\leq100$. This is rather easy.

Then, introduce $z$ in the play.