Let $n < 10$ be a non-negative integer. How many integers from $0$ to $999$ inclusive have the sum of their digits equal to $n$? Give your answer in terms of $n$. Hint: Try first for integers from $0$ to $99$.
1. How many $3$ digit numbers, whose digits consist solely of even numbers, exist?
2. A ternary number consists of only $2$s, $1$s and $0$s. How many values can be represented by a $7$ digit ternary number?

You can use inspection (there are only 10 cases) for the case of max 2 digits to get an expression in terms of $n$.

For the case of max 3 digits, what if you fix one digit to some value $p\le n$?

What must be the sum of the remaining max 2 digits in terms of $p$ and $n$?

How many max 2 digits numbers have their digit sum equal to $n-p$?

How should we incorporate that for all values of $p$?

For max $99$ (max 2 digits) case one can observe, by inspection, that there are $n+1$ numbers whose digit sum is $n$.

When there are max 3 digits, let’s fix one of the digits to $p\le n$. The sum of the remaining max 2 digits must thus equal $n-p,$ and we know there are $n-p+1$ numbers with that property. Taking all possible values of $p$ we get \begin{aligned}\sum_{p=0}^n (n-p+1)&=(n+1)\sum_{p=0}^n 1-\sum_{p=0}^n p\\&=\frac{(n+1)(n+2)}{2}.\end{aligned}