Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
up vote 2 down vote accepted

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

share|cite|improve this answer
we can't use 0 at first place as it will make a 4 digit number. 0 can be used anywhere else. – Bharat Kul Ratan Oct 13 '12 at 12:27
If we want $0$, you'd have to change the $9$-s in my answer to $10$-s. If we pick $0$, it would have to be in the rightmost position anyways, as having the zero in any other position would make the rest of the digits impossible to place. So $0$ gets no special treatment, and the answer is $\binom{10}{5} = 252$ – Arthur Oct 13 '12 at 12:56
thanks, got it. I don't know why I treated this as a tough problem :( – Bharat Kul Ratan Oct 13 '12 at 13:43
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. – Arthur Oct 13 '12 at 13:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.