2012 AMC 12B Problem 18

Once $a_1=k$ is fixed, the numbers $k,k+1,\ldots,10$ must appear left to right in increasing order, and the numbers $1,\ldots,k-1$ must appear from right to left in increasing order (so each new small number has its successor already placed). For each $k,$ the list is determined by choosing which of the $9$ positions after the first hold the numbers below $k,$ giving $\binom{9}{k-1}$ lists. Summing, $$\begin{aligned} \sum_{k=1}^{10}\binom{9}{k-1} &= \sum_{j=0}^{9}\binom{9}{j} \\ &= 2^9=512. \end{aligned}$$ Thus, the correct answer is B.