AMC10 2012 B

Answer (B): If $a_1=1$, then the list must be an increasing sequence. Otherwise let $k=a_1$. Then the numbers 1 through $k-1$ must appear in increasing order from right to left, and the numbers from $k$ through 10 must appear in increasing order from left to right. For $2\le k\le 10$ there are $\binom{9}{k-1}$ ways to choose positions in the list for the numbers from 1 through $k-1$, and the positions of the remaining numbers are then determined. The number of lists is therefore $$ 1+\sum_{k=2}^{10}\binom{9}{k-1}=\sum_{k=0}^{9}\binom{9}{k}=2^9=512. $$ OR If $a_{10}$ is not 1 or 10, then numbers larger than $a_{10}$ must appear in reverse order in the list, and numbers smaller than $a_{10}$ must appear in order. However, 1 and 10 cannot both appear first in the list, so the placement of either 1 or 10 would violate the given conditions. Hence $a_{10}=1$ or 10. By similar reasoning, when reading the list from right to left the number that appears next must be the smallest or largest unused integer from 1 to 10. This gives 2 choices for each term until there is one number left. Hence there are $2^9=512$ choices.