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How many four-digit positive integers are there such that all digits are different?

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  • 2
    $\begingroup$ @Shepard I can tell you're busy playing Mass Effects ... but at least can you show some of your work? or whatever you have attempted? that way people would be more likely, and willing to help you $\endgroup$ – user67258 Aug 3 '13 at 4:24
  • $\begingroup$ We should play and talk about math at the same time! $\endgroup$ – Commander Shepard Aug 3 '13 at 4:57
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For a valid $4$-digit number, the Highest Significant digit will be among $1\cdots9$ so can have $9$ values

From the next onwards, they can assume $0\cdots 9$ excluding the ones already selected

So, the number of combinations will be $$9\cdot9\cdot8\cdot7$$

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Hint:

The first digit can be any number from $1$ to $9$. The next three digits can be any number from $0$ to $9$. If we choose each digit one at a time and stipulate that each digit is different from the previous choices, how many choices do we have for each digit? Multiply these together to get your answer.

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Allowing for a zero first digit

10 * 9 * 8 * 7

10 possible options for the first digit, 9 remaining options for the second digit, 8 remaining options for the third digit ...

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    $\begingroup$ allowing for a zero first digit would make it a 3-digit number $\endgroup$ – user67258 Aug 3 '13 at 4:26

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