$36 \cdot 12$ If the 120 permutations of $A H S M E$ are considered as ordinary five-letter words and are arranged in alphabetical order, the last letter of the 86th word is
(A) A.
(B) $H$.
(C) $S$.
(D) $M$.
(E) $E$.
(37th American High School Mathematics Examination, 1986)
[Solution] It is clear that this is equivalent to finding: the last digit of the 86th number in the sequence of five-digit numbers formed by the digits $1,2,3,4,5$ without repetition, arranged in ascending order.
Among the five-digit numbers formed by these five digits, those starting with 1, 2, or 3 are
$$
C_{3}^{1} \cdot P_{4}=72 \text{ in total. }
$$

Next, among the five-digit numbers starting with 4, those with 1 or 2 as the thousandth digit are
$$
C_{2}^{1} P_{3}=12 \text{ in total. }
$$

At this point, there are already $72+12=84$ numbers. The next should be the five-digit numbers starting with 4 and having 3 as the thousandth digit: $43125, 43152, \cdots$
43152 is the 86th number, and its last digit is 2.
Note the correspondence: $A \leftrightarrow 1, \quad E \leftrightarrow 2, \quad H \leftrightarrow 3, \quad M \leftrightarrow 4, \quad S \leftrightarrow 5$. Therefore, the answer is $(E)$.