Digits with sum of 9

To start this off, write down the equation: a + b + c = 9.

Consider this: a + b + c = 9 and each digit is greater than or equal to 0.

If you split this problem into cases, it will be harder to solve.

Add 1 to each digit: a + 1 + b + 1 + c + 1 = 12 so that each digit will be greater than 0.

Then, by using the stars and bars method, you can visualize 12 stars with 11 gaps in between them.

To find out how many possibilities there are, use 11C2, since you use two slices to have three groups, and you will get 55.

Now consider that a = 0:

Since a is already 0 => b + c = 9 and b and c is greater than or equal to 0. Using the previous method, add 2 to b and c to get the sum to 11. There are 11 stars and 10 gaps. 10C1 = 10.

Eliminate the situation where a = 0: 55 - 10 = 45. There are $\boxed{45}$ different combinations of the 3 digits.

"sum of the digits is $9$ " immediately suggests looking at multiples of $9$ .

The trick is to notice that for every pair of positive multiples of $9$ that sum to $999$ , one has digit sum $\textbf9$ and one has digit sum $18$ .

These pairs are $(\textbf9,990),(\textbf{18},981),\ldots,(99,\textbf{900}),(\textbf{108},891), (\textbf{117},882),\ldots,(495,\textbf{504})$ . Note that $10$ of the numbers with digit sum $9$ are less than $100$ , so don't count towards the answer.

Counting up, there are $\frac{495}{9}-10=\boxed{45}$ such three digit numbers.