Let ($a_1$, $a_2$, ... $a_{10}$) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?

$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880$
If we have 1 as the first number, then the only possible list is $(1,2,3,4,5,6,7,8,9,10)$. 
If we have 2 as the first number, then we have 9 ways to choose where the $1$ goes, and the numbers ascend from the first number, $2$, with the exception of the $1$.
For example, $(2,3,1,4,5,6,7,8,9,10)$, or $(2,3,4,1,5,6,7,8,9,10)$. There are $\dbinom{9}{1}$ ways to do so.
If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are $\dbinom{9}{2}$ ways to do this.
In the same way, the total number of lists is:
$\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}$
By the [binomial theorem](https://artofproblemsolving.com/wiki/index.php/Binomial_theorem), this is $2^{9}$ = $512$, or $\boxed{\textbf{(B)}}$