# Three digit numbers where two digits are the same — I'm wrong but where?

1. there are $9 * 9 * 8$ three digit numbers where all digits are different so $1000 - 9 * 9 * 8$ which have at least two digits that are the same and then there are $9$ numbers which have exactly three same digits so the number of three digit numbers where exactly two digits are the same is $1000 - 9 * 9 * 8 - 9 = 343$.

2. Let's pick a digit between $1-9$. If we decide a three digit number should contain this digit twice then there are three positions for the remaining digit, when it's the first digit can be 8 kinds (144, 244, 344, 544 etc), the second and the three $9-9$ each (404, 414 etc) so far we have $9 * ( 8 + 9 + 9 )$ . Finally if the repeating digit is $0$ then the first digit can be any of $1-9$ so the end result is $9 * ( 9 + 9 + 9) = 243$.

Both can't be right. So which one is wrong and why?

• There are $900$ three digit numbers, not $1000$, so there are $900-9\cdot 9\cdot 8$ three digit numbers that have at least two equal digits, not $1000-9\cdot 9\cdot 8$. – user236182 May 26 '16 at 17:53

In your first point, there are $900$ three digit numbers, not $1000$. This gives the missing $100$.