For how many integers having between 1 and 10 digits (base 10) are all of their digits when read from left to right monotonically increasing? In other words, every digit is less than or equal to all of those to its right. For example, 244467889 is one of them, and 0 is another, but there are more.

C(19,10), which simplifies to 19*17*13*11*2 = 92,378. This is the same number arrived at by Charlie and Jer, although my methods were different.

It turns out that this is another problem of items with separators.

Consider 9 items separated by 10 separators. Interpret each separator as a digit, whose value equals the number of items (not counting separators) to its left.

Then |oo|oo|||oo|o|o||o| = 0244467889
||||||||oooooooo||o = 0000000088

Every monotonous integer less than 10^10 corresponds to one arrangement of 9 items separated by 10 separators, and every arrangement of 9 items separated by 10 separators corresponds to a distinct monotonous integer less than 10^10.

So, the total number of monotonous integers between 0 and 10^m = C(9+m,m).
The right hand side of Charlie's table shows this series: 10, 55, 715, etc.

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In case anybody is interested, this is an original problem. I have an RSA security token that generates a new 8 digit integer every minute, and last week it generated an integer whose digits increased monotonically. Which set me to wondering. I made up the term "monotonous" to describe this integer; it is not to the best of my knowledge accepted mathematical terminology.

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