I tried to find all the numbers between 100 and 999, that consist of (pairwise) different ciphers. So the first would be 102 and the last would be 987.
I think there are 9*9*8 such numbers, here's the argumentation I'm not sure about:
The first digit contains only 1-9, and for the second digit there's one possibility less. But as zero is now allowed again, it remains 9. And for the last digit again one possibility less.