Let $(a_1,a_2, \dots ,a_{10})$ be a list of the first 10 positive integers such that for each $2 \le i \le 10$ either $a_i+1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ 362,880$
Let $1\leq k\leq 10$. Assume that $a_1=k$. If $k1$, the first number appear after $k$ that is less than $k$ must be $k-1$, and then $k-2$, and so forth.
To count the number of possibilities when $a_1=k$ is given, we set up $9$ spots after $k$, and assign $k-1$ of them to the numbers less than $k$ and the rest to the numbers greater than $k$. The number of ways in doing so is $9$ choose $k-1$.
Therefore, when summing up the cases from $k=1$ to $10$, we get
\[\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} = 2^9=512 ...... \framebox{B}\]