How many four digit numbers in which each digit is at least as large as the previous digit? Given that the first digit has to be between 1 and 9, each subsequent digit depends on the previous. If they were strictly increasing, then it would be $\binom{9}{4}$ numbers. My idea is that for each digit after the first, there is an additional choice compared to the strictly increasing numbers, but I'm not sure how to translate that into an expression. Any help is appreciated!

Hint:
If $$1\leq a_1<a_2<a_3<a_4\leq 12$$ then $$1\leq a_1\leq a_2-1\leq a_3-2\leq a_4-3\leq  9,$$
and visa versa.
