Number of integer solutions of $x_1+x_2+⋯+x_n=m$ when $x_i$ can be negative, simpler solution

Find the number of integer solutions of $x_1+x_2+x_3+x_4=30$ when $x_i \leq20$.

Since $x_i \geq -30$, I defined $y_i=x_i+30$, $0 \leq y_i \leq 50$, and we get $y_1+...+y_4=150$, and by inclusion-exclusion we get $\binom{153}{3}-4\binom{102}{3}+\binom{4}{2}\binom{51}{3}=\boxed {\binom {53} {3}}$

So the answer is ${\binom {53} 3}$, which suggests there's a clever substitution that'll make this into the classic $\binom{n+k-1}{k}$ case, of $\sum_{i=1}^n x_i=k$.

I have found numerous variants of this question here, but not this one.

This come from the fact that if $$x_1+x_2+x_3+x_4=30$$ with $$x_i\le 20$$
Then, let $z_i=20-x_i$ $$z_1+z_2+z_3+z_4=50$$
with $$0\le z_i$$
So $(z_i)$ is a $4$-partition of $50$, there is $\binom{50+4-1}{4-1}$ such partitions.