I don't know how to solve for the number of integer non negative solutions of following system: $$ \left\{\begin{array}{l}{x_{3}+x_{4}+x_{5}=6} \\ {x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=27}\end{array}\right. $$ I do know how to solve them separately but not when they are joint.
Advancements(I):
I realized later that I could look at this as if I was to solve another system of equations: $$ \left\{\begin{array}{l}{x_{3}+x_{4}+x_{5}=6} \\ {x_{1}+x_{2}=21}\end{array}\right. $$
The number of solutions for each equation are ${8 \choose 2}$ and ${22 \choose 1}$ respectively. So would the solution be ${8 \choose 2}+{22 \choose 1}=28+22=50$?
Advancements(II):
The previous reasoning is wrong, I think this is it:
I obtain the solutions for $x_{3}+x_{4}+x_{5}=6$ which is ${8 \choose 2}= 28.$
Since this equations determines the second, these are the solutions for the system.
You have correctly found that the "advanced" equations have $\binom82$ and $\binom{22}1$ solutions, respectively. Since you can combine any triple which is a solution to the first equation with any pair which is a solution to the second equation the overall number of solutions is $$\binom82\binom{22}1.$$