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.

@ Jer: The sum for C(8+m,8) equals for (m from 0 to m)

C(9+m,m).

For m=10 equals C(19,9)

**i.e, 92,378**

I have found the **1 to 1 mapping** of all 10 digit numbers from 0 to 9,999,999,9999 into a binary string of 19 bits;( 10 zeroes and 9 ones.)

After 2 days of "suffering", I have on my hands a 10 minute solution, **will describe later.**

*Edited on ***December 31, 2015, 3:26 am**