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 ?
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 ?
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 $\binom{9}{N} = \frac{9!}{N!(9-N!)}$.