$n < 10$
be a nonnegative 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 $99$
.

 How many
$3$
digit numbers, whose digits consist solely of even numbers, exist?  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?
 How many

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
$np$
? 
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$np,$
and we know there are$np+1$
numbers with that property. Taking all possible values of$p$
we get$$\begin{aligned}\sum_{p=0}^n (np+1)&=(n+1)\sum_{p=0}^n 1\sum_{p=0}^n p\\&=\frac{(n+1)(n+2)}{2}.\end{aligned}$$