yes that approach works for this problem.

9 is possibility of first digit
8 is possibility of second digit
only 1 possiblity for last digit.

Times 3 because there are 3 digits

So if we knew that the 1st and 3rd digits had to be the same, it would just be 9*8*1? Or would it be 9*8*1 + 1*8*9 ?

your answer is perfect....

3 digits, can be filled by 2 similar digits in 3 ways, 2 similar digits (barring 0) can be picked in 9 ways...and third digit can be picked in 8 ways

3*9*8 = 216

Let us include 0 and then solve it....

EricKryk wrote:So if we knew that the 1st and 3rd digits had to be the same, it would just be 9*8*1? Or would it be 9*8*1 + 1*8*9 ?

It would be just 9*8*1.

Let me explain the problem and the solution.
Let us take an example. let us assume that repeated digit is 3 and other digit is 7. So we can make the following combinations.

337, 373, 733 => 3 variations.

The solution is similar to what everyone have solved above.
Lets start from Hundreds position and move to units position.

Hundred's position can be filled in 9 ways coz we have 9 digits (1 - 9).
Ten's position can be filled in 1 way only (coz we want to have the digit in
hundred's position repeated).
units position can be filled in 8 ways coz we dont want the digit that we used for hundred's and ten's positions.

ie. 9*1*8 possibilities for 1 variation.
So for 3 variations we have 9*1*8*3 = 216

Ht helps
-V