We divide into cases. The only possible values for $x_{4}$ are $x_{4}= 0,1,2,3$, since otherwise the left hand side is too large.
- Case $x_{4}=0:$ We get $3\left(x_{1}+x_{2}+x_{3}\right)=22$. No solutions as the right hand side is not a multiple of $3$
- Case $x_{4}=1:$ We get $3\left(x_{1}+x_{2}+x_{3}\right)=15$, so $x_{1}+x_{2}+x_{3}=5$ which has $\left(\begin{array}{l}7 \\ 2\end{array}\right)=21$ solutions.
- Case $x_{4}=2:$ We get $3\left(x_{1}+x_{2}+x_{3}\right)=8$, no solutions.
- Case $x_{4}=3:$ We get $3\left(x_{1}+x_{2}+x_{3}\right)=1$, no solutions.
Therefore, there are $21$ solutions in total.
https://www.goclasses.in/