In my probability book I learned how to solve this kind of problem: how many vectors of the form $(i\_1,i\_2,...,i\_r)$ exist that the sum of its components is $n$ and $i\_k>0$ for all $k$? The answer uses the stars and bars argument and is $n-1\choose r-1$.

OK, now I have this really simple problem: imagine you'll roll $2$ six-faced dice, how many ways there are in which the sum of them is $n$? 

First I thought to use the before mentioned method, considering a result as a vector of the form $(i\_1,i\_2)$ but I realized that I have a new restriction, $i\_1,i\_2 \leq 6$! I thought about it and I don't know how to compute it without considering case by case ($n=2,n=3,...,n=12$), I've done it with the six-faced die, since there's only $11$ cases and found a nice 'triangular pattern', but what if the dice had each $20$ faces? Is there a more smart way to calculate it?