If ${x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}\le 20$ where ${x}_{1},{x}_{2},{x}_{3},{x}_{4}$ are non negative integers , then how many possible outcome are there ?

It is basic counting problem, we solve it by adding an extra non-negative integer such that ${x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}=20$, and then use star and bars . Hence the answer is $(}\genfrac{}{}{0ex}{}{20+5-1}{20}{\textstyle )$.

However , there is something that stuck in my mind such that why are we adding only one extra variable ${x}_{5}$, to be clearer , why dont we add more than one extra variable. For example , why did not i find the solution of ${x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}+{x}_{6}=20$ where the extra terms ${x}_{5}$ and ${x}_{6}$ are non-negative, too. Briefly, why did we restrict ourselved with only one extra variable? Can you enlighten me ?