The number of ordered triples (a,b,c) of positive integers with a+b+c=20 and each of a,b,c≤ 9 is

Without the upper bound, the number of positive solutions of $a+b+c=20$ is $\binom{19}{2}=171$. Substitute $a'=a-1$ etc. so the variables are non-negative with sum $17$ and each at most $8$. Count the violations by inclusion-exclusion. For a violation $a\ge10$, set $a=10+a''$ with $a''\ge0$ and $b,c\ge1$: then $a''+(b-1)+(c-1)=8$ with all parts non-negative, so the number of solutions is $\binom{10}{2}=45$. There are 3 such variables, giving $135$. Two variables cannot both be $\ge10$ because $10+10+1>20$. So the count is $171-135=36$. A direct enumeration over $a,b$ with $1\le20-a-b\le9$ and $a,b\le9$ confirms $36$.