Counting Problems with Digits

"How many 3-digit numbers have digits summing to $n$?" is a stars-and-bars count with a constraint: the leading digit cannot be zero and no digit exceeds 9.

Substitute $a = a' + 1$ so the leading digit starts at zero, then count solutions of $a' + b + c = n - 1$ with each at most 9. Without the upper bound that is $\binom{n+1}{2}$; subtract the cases where one variable exceeds 9.

For small $n$ the bound never bites, so the answer is just $\binom{n+1}{2}$.