Firstly, this expression can be rewritten as $$\sum_{m=0}^{2009}\sum_{n=0}^{m}\binom{2009}{m}\binom{m}{n}=\sum_{m=0}^{2009}\binom{2009}{m}\sum_{n=0}^{m}\binom{m}{n}$$

The inner part is the sum of coefficients in $(x+y)^m$ which can be obtained by setting $x=y=1$. Therefore, $$\sum_{m=0}^{2009}\binom{2009}{m}\sum_{n=0}^{m}\binom{m}{n}=\sum_{m=0}^{2009}\binom{2009}{m}(1+1)^m = \sum_{m=0}^{2009}\binom{2009}{m}2^m$$

Now this expression is the sum of all the coefficients of $(x+2)^{2009}$ which can be obtained by setting $x=1$. This means that the final answer is $$(1+2)^{2009}=\boxed{3^{2009}}$$