combinations [fundemental counting principal]

Question

How many 3-digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero?

I have approached the following problem like this.

1. We know that zero is not included so we can choose anything from 1-9 which gives us 9 options.
2. We have to select a 3 digit integer therefore we have _ _ _ 3 different stages and we can use the fundamental counting principle.
3. We have 1 constraint that is that none of the digits can be repeated more then 2 times.

So i did 9 * 9 * 9 and got 729. This is wrong though because i have not taken into account the constrain of not having 3 repeated digits.

Can anyone explain how i would remove the repeated digits from this?

Answer 1

From the $9^3$ possible three-digit numbers with no zeros, subtract the number of ways that digits can be repeated three times. There are 9 numbers, each of which can be repeated three times in only one way, thus forming exactly 9 possible unique three-digit numbers that have three repeating digits:

You should obtain: $$9^3 - 9 = 720 \;\;\text{combinations}$$