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There is a problem in which I am obtaining a different answer than my professor. The problem is as follows:

How many integers between $1$ and $1000$ use exactly three digits?

The professor shows the solution as: $9 \cdot 9 \cdot 8=648$, but I have no idea where those numbers are coming from. On the other hand, I say that, excluding $1$ to $99$ and $1000$, there are $900$ integers that use exactly three digits. Can someone please explain which one of us is right and why?

Thank you!

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    $\begingroup$ Your list would include $111$, which uses exactly one digit - not three. The wording is a bit ambiguous, but I'd bet this is what's intended. $\endgroup$
    – user296602
    Apr 12, 2016 at 2:45

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Your professor is not allowing integers where two digits match- s/he wants three distinct digits. The word distinct is not in the problem as quoted. To use exactly three digits, the number needs to not have a leading zero, so be greater than $99$. This means you have $9$ choices for the first digit. Then the second digit cannot match the first, so you have $9$ choices for it. Then the third ...

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  • $\begingroup$ Okay, thank you! I think I understand now. $\endgroup$ Apr 12, 2016 at 2:47
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Professor has shown the result for all those numbers in which digits are not repeated.

You are in the need of three digits number in which digits are non repeating, so let the structure of number be “_ _ _"

$•$ now except “0" any digits can be placed at $100^{th}$ place,so it can be filled in 9 ways

$•$ similarly $ten^{th} $place can be filled in 9 ways (including “0" and excluding number used in $ 100^{th}$ place)

$•$ once place can be filled in 8 ways (excluding number used in $100^{th}$ ,$10^{th}$ place )

So total number of ways equal to $9×9×8=\boxed{648}$

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