When a die is rolled, one of the first six positive integers is obtained. Suppose that the die is rolled five times and the sum of the five integers thus obtained is added. The five throws constitute a trial. Find the number of possible trials such that the sum is at most $12$.
This can be written as $x_1+x_2+x_3+x_4+x_5\leq 12$ where $1\leq x_i\leq6$ for $i=1,2,...,5$. However, this restriction is making this problem difficult for me. I know that I can figure out the solution by considering the generating function $G(x)=(x+x^2+x^3+x^4+x^5+x^6)^5$ and take the sum of all the coefficients that are less than or equal to $12$, but I want to know how to do this with the stars and bars technique.
I know that I can rewrite the linear inequality as $x_1'+x_2'+x_3'+x_4'+x_5'\leq 7$ so we can count the number of positive integral solutions, but then we have still over counted and we need to use the restriction $x_i\leq 6$. I do not know how to do this. Any solutions or hints are greatly appreciated.