# 1986 AHSME Problems/Problem 10

## Problem

The $120$ permutations of the $AHSME$ are arranged in dictionary order as if each were an ordinary five-letter word. The last letter of the $85$th word in this list is: $\textbf{(A)}\ \text{A} \qquad \textbf{(B)}\ \text{H} \qquad \textbf{(C)}\ \text{S} \qquad \textbf{(D)}\ \text{M}\qquad \textbf{(E)}\ \text{E}$

## Solution 1

We could list out all of the possible combinations in dictionary order. $73rd:$ MAEHS $74th:$ MAESH $75th:$ MAHES $76th:$ MAHSE $77th:$ MASEH $78th:$ MASHE $79th:$ MEAHS $80th:$ MEASH $81th:$ MEHAS $82th:$ MEHSA $83th:$ MESAH $84th:$ MESHA $85th:$ MHAES

We find that the 85th combination ends with the letter S. So the answer is $\textbf{(C)}\ S$.

## Solution 2

We can do this problem without having to list out every single combination. There are $5$ distinct letters, so therefore there are $5!=120$ ways to rearrange the letters. We can divide the $120$ different combinations into 5 groups. Words that start with $A$, words that start with $E$ and so on... Combinations $1$- $24$ start with $A$, combinations $25$- $48$ start with $E$, combinations $49$- $72$ start with $H$, combinations $73$- $96$ start with $M$, and combinations $97$- $120$ start with $S$. We are only concerned with combination $85$, so we focus on combinations $73$- $96$. We can divide the remaining 24 combinations into 4 groups of 6, based upon the second letter. Combinations $73$- $78$ begin with $MA$, combinations $79$- $84$ begin with $ME$, combinations $85$- $90$ begin with $MH$, and combinations $91$- $96$ begin with $MS$. Combination $85$ begins with $MH$. Now we can fill in the rest of the letters in alphabetical order and get $MHAES$. The last letter of the word is $S$, so the answer is $\textbf{(C)}\ S$ .

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 