Probability of the digits in a number being non-descending If I choose a random $16$ digit number, what is the probability that each digit is greater-than or equal-to the previous? Leading 0's are required; numbers are in the range $[0000000000000000-9999999999999999]$.
The total number of possibilities is $10^{16}$.
As the result of a simple recursive program, I believe the number of desirable results is $2042975$, making the probability $0.00000002042975\%$
However, I cannot come up with the mathematics behind it.
Something interesting to note; if I'm doing the same for a random 2 digit number, the solution is $$10+9+8+7+6+5+4+3+2+1 = 55$$
 A: The number of ways of creating a qualifying number is $${25  \choose 9}=2042975$$ 
as you found.
The reasoning is that you can increment $9$ times - possibly before and after the $16$ digits - from $0$ to $9$. Choosing the location of those increments amongst the $16$ digits gives the value above.
For example, the following increment locations ($+$) amongst the digits ($N$):
$$++NN+NNN+N+NN+NNNN+NNN+N+$$
gives the number 
$$2233345566667778$$
Any number that consists of digits that are always greater than or equal to the previous (apologies for confusion, my first framed answer was for subsequent) can be represented like this.
A: Just another way of explaining: 
Imagine a $10$ sided die being thrown $16$ times, and tallies of $0$ thru $9$ noted.
A tally of $\;\;\fbox{2}\fbox{3}\fbox{0}\fbox{2}\fbox{3}\fbox{1}\fbox{2}\fbox{2}\fbox{0}\fbox{1}\;$, e.g.would mean the number $0011133444566779$
Using stars and bars,$\;\binom{16+10-1}{10-1}$ non-descending tallies are possible,
and dividing by $10^{16}$ will yield the desired probability
