jmc

Clearly, $a_6=1$. Now, consider selecting $5$ of the remaining $11$ values. Sort these values in descending order, and sort the other $6$ values in ascending order. Now, let the $5$ selected values be $a_1$ through $a_5$, and let the remaining $6$ be $a_7$ through $a_{12}$. It is now clear that there is a bijection between the number of ways to select $5$ values from $11$ and ordered 12-tuples $(a_1,\dots ,a_{12})$. Thus, there will be $\left(\genfrac{}{}{0ex}{}{11}{5}\right)=\boxed{462}$ such ordered 12-tuples.