10. Consider the 12C permutations of $A H S M E$ as ordinary five-letter words and arrange them in dictionary order. The last letter of the 86th word is
(A) $A$; (B) $H$; (C) $S$; (D) $M$; (E) $E$.
$\frac{10}{\mathrm{E}}$
10. The first $4!=24$ words start with the letter $A$, the next 24 words start with $E$, and the following 24 words start with $H$. Therefore, the 86th word starts with $M$. And it is the $86-72=14$th word of such: $\quad / M$ starts with the first 6 words starting with $M A$, the next few words start with $M E$, and the word we are looking for starts with $M H$. The first word starting with $M H$ is $M H A E S$, and the second word is $M H A S E$, so $E$ is the letter we are looking for.