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How many integers from 1-999 do not have any repeated digits?

The answer is explained in this link, but why is the last set 9*9*8? Why not 9*9*9?

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In the link, they claim that $9\cdot 9\cdot 8$ gives the number of integers from $100$ to $999$ with no repeated digits. You have three digits $d_1d_2d_3$ to work with. Begin with the first digit $d_1$: it can be any number in $\{1, \ldots, 9\}$. When choosing the second digit $d_2$ you must avoid repeating $d_1$, but you can now use $0$, so you select from $\{0, \ldots, 9\} - \{d_1\}$. So while the first two factors of $9 \cdot 9 \cdot 8$ are the same, there is an interesting difference between picking $d_1$ and picking $d_2$. Similarly, the the last digit must come from $\{0, \ldots, 9\} - \{d_1, d_2\}$, and this set has size $8$.

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Thank you so much, Dylan! I had been attempting this problem for at least an hour! – user754950 Dec 8 '11 at 2:00
@user754950 No problem. Hope it makes sense now. – Dylan Moreland Dec 9 '11 at 20:39

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