Find the number of three-digit numbers (integers from 100 to 999) that contain no two consecutive equal digits.
My thoughts were, there a 1-9 choices for position one; so 9 choices. For the second position there are 0-9 choices but it cannot be the same as the first, so there are 9 choices, and for the last position there are 0-9 choices, but it can't be the same as the second so there are again 9 choices. So my answer is $9^3 = 729$.
I have checked on this website and found a similar problem, but their reasoning and answer did not match with my answer. Please tell me what I am doing wrong and what the correct answer is.