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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.

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