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

You have to count the number of integers in two situations:

1. all three digits are different. In this case you have a permutation of 3 out of 9 which is 9*8*7=504 integers
2. one digit appears twice. In this case you first choose 2 digits from 9 which is 9*8/2=36 ways. and each time you have 3 ways to permute. For example if you choose 1 and 2 you can get 112,121,or 211 three integers. So in total you have 3*36=108 integers.

add 1 and 2, 504+108=612.