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How many ways to write a 5 digit number so that every digit is scritctly greater than the digit on it's right ?

How could we derive a formula for such a N digit number where N <= 9 ?

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This is the same as picking $5$ different digits out of the nine possible (I assume you don't want $0$ involved), and then ordering them afterwards. So $\binom{9}{5} = 126$. For $N$ digits the general answer is $\binom{9}{N} = \frac{9!}{N!(9-N!)}$.

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    It's a matter of perspective. The moment you start to tunnel in on the set of integers between $10\:000$ and $99\:999$, and try to classify and count the right ones, it becomes a hard problem.2012-10-13