Find number of integral solutions for $x+y+z= -20$ such that $x,y,z$ are negative integral multiples of $4$?

Question

I tried solving this but couldn't find the right approach,I can find it manually as only few options are viable, but I wanted to understand the method.

For eg: We can easily see that only few values like

-4,-4,-12 -8,-8,-4,

So,6 ways but wanted to understand the method involved.

Answer 1

Write $x = -4a$ , $y = -4b$ , $z = -4c$ when $a$,$b$,$c$ is a positive integers.

Back in the problem statement , we get $ a + b + c = 5$

Let $a$ = $a_1 + 1$ , $b$ = $b_1 + 1$ and $c$ = $c_1 + 1$

So, $a_1 + b_1 + c_1 = 2$

By Stars & Bars , number of solution is (2 + 3 - 1) choose 2.

Hence , number of solutions is 6.

(Number of solutions $x_1$ + $x_2$ + $x_3 = r$ when all $x_1$ , $x_2$ , $x_3$ is non-negative integers is like numbers of ways to put r identical balls into 3 boxes. Which is ($r + 3 - 1$) choose ($r$) by stars and bars).