Don't get smaller- Get bigger!

Note that there is a bijection between sets of 6 digits, allowing repetition, and 6 digit numbers satisfying the condition, because once the digits are chosen, there is exactly one way to place the digits in ascending order. Thus we need to count the number of sets of 6 digits, where only the number of appearances of each digit is significant.

Let $x_1$ be the number of 1's, $x_2$ be the number of 2's, and so on. The equation we now have is $x_1+x_2+...+x_9=6$ , where $x_i$ is a nonnegative integer for all $1\le i\le9$ . By a "stars-and-bars" argument, the number of such 9-tuples $(x_1, x_2, ... , x_9)$ is $\dbinom{m+n-1}{n-1}=\dbinom{9+6-1}{6}=\dbinom{14}{6}=\boxed{3003}$ .