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# Digits on the rise

### #1

Posted 12 October 2013 - 02:36 AM

What do 15 489 1256 and 24578 have in common?

They are positive integers whose digits are strictly increasing left to right.

How many of them are there?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #2

Posted 12 October 2013 - 02:55 AM

Please clarify if you mean

- how many positive increasing-digit integers in {the set of all possible digit combinations in the stated problem} or
- how many positive increasing-digit integers in {the set of all positive integers}

(I have a feeling it's the latter option...)

**Edited by Dariusray, 12 October 2013 - 02:55 AM.**

### #3

Posted 12 October 2013 - 03:24 AM

Correct. The latter: positive integers whose digits are strictly increasing left to right.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #4

Posted 12 October 2013 - 05:30 AM

### #5

Posted 12 October 2013 - 05:35 AM Best Answer

### #6

Posted 12 October 2013 - 10:21 AM

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #7

Posted 13 October 2013 - 03:09 AM

There's an even simpler observation that immediately gives the correct answer.

I assumed there is,

### #8

Posted 13 October 2013 - 12:38 PM

There's an even simpler observation that immediately gives the correct answer.

I assumed there is,Spoiler for given the fact that the answer is, but I'm not sure right now why that is.

The power set (the set of all subsets) of a set of n objects has size 2

^{n}. For 9 digits, n=9. Then subtract off the empty set.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

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