How many Ascending Numbers there are?

No ascending number with $N$ digits can contain the digit $0$ , since $0$ is the smallest of the digits, and this would mean that the ascending number would then begin with $0$ , which would make it a number with fewer than $N$ digits. Thus ascending numbers can only contain the digits $1,2,3,4,5,6,7,8,9$ . The number of ascending numbers with $N$ digits is therefore the number of $9$ -tuples $(a_1,a_2,...,a_9)$ of integers such that $a_1,a_2,...,a_9 \ge 0$ and $a_1 + a_2 + \cdots + a_9 = N$ (so that the ascending number consists of $a_1$ copies of $1$ , $a_2$ copies of $2$ , and so on, up to $a_9$ copies of $9$ , written in that order.

Standard combinatorics tells us that the number of ascending numbers with $N$ digits is $p_N \; = \; \binom{8+N}{N}$ Thus we want to find $p_{10^7} \pmod{10^9}$ . Since $\begin{aligned} p_{10^7} & = \; \binom{10^7+8}{10^7} \\ & = \; 2480167658743700408075402393106827480451696455750001 \end{aligned}$ this makes the answer $\boxed{455750001}$ .

We consider the continuous digits which share the same number one block. there are at least 1 block and at most 9 block in the sequence. Suppose there are $n$ digits and $m$ blocks. for block=1, we can fill in 9 numbers, and there's no "paddle" to break it into several blocks.

i.e. $\binom{9}{1}*\binom{n-1}{0}$

for block=2, we can fill in 9 numbers in 2 blocks, and there's one "paddle" to break it into 2 parts, which can be put into $n-1$ places.

i.e. $\binom{9}{2}*\binom{n-1}{1}$

by the same token, for block= $m$ $(1 \leq m \leq 9 )$ , we can fill in 9 numbers in $m$ blocks, and there's $m-1$ "paddles" in $n-1$ places to break into $m$ parts.

i.e. $\binom{9}{m}*\binom{n-1}{m-1}$

so the answer is:

$\sum_{m=1}^9 \binom{9}{m}*\binom{n-1}{m-1}$ $mod 10^9$