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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.

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Yes, your argument is correct. – Ilmari Karonen Aug 2 '14 at 23:12
up vote 5 down vote accepted

Well, you can pick $3$ out of $10$ numbers, so $10\choose 3$. For each of those selections, you can order them in $3!$ ways. You just have to rule out the numbers that start with a $0$. You can use a similar argument to get to $2!·{9\choose 2}$.

So the total number is $\frac{10!}{7!} - \frac{9!}{7!} = 648$

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Nice approach ! – 355durch113 Aug 2 '14 at 22:41

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