2002 AMC 10B Problem 9

Alphabetically the letters are $A < M < O < S < U$. Count how many words are listed before $USAMO$.

First letter earlier than $U$: any of $A, M, O, S$, followed by any arrangement of the other four letters. That is $4 \cdot 4! = 96$ words.

First letter $U$, second letter earlier than $S$: the second letter is $A$, $M$ or $O$, then any arrangement of the remaining three. That is $3 \cdot 3! = 18$ words.

First two letters $US$: the remaining letters $A, M, O$ in alphabetical order spell $AMO$, so $USAMO$ is the first word in this block.

So $96 + 18 = 114$ words precede $USAMO$, and it sits in position $115$.

The answer is $\boxed{\textbf{(D)}\ 115}$.

Alphabetical order is a nested sort: all words with a smaller first letter come first, then within a tie, all with a smaller second letter, and so on. Counting the words that precede a target is therefore a sum of blocks, each block being (number of smaller letters available at that slot) times (arrangements of the letters left). Add $1$ at the end to convert "words before" into "position."

Count from the back. There are $5! = 120$ words, and after $USAMO$ come exactly the $US$-words with the tail bigger than $AMO$: $USAOM, USMAO, USMOA, USOAM, USOMA$, five of them. So $USAMO$ is number $120 - 5 = 115$.

• Reporting 114, the number of words that come before USAMO, instead of its position 115; or sorting the letters in the order U, S, A, M, O rather than alphabetically.
• Forgetting the second block (the $18$ words $UA\ldots$, $UM\ldots$, $UO\ldots$) and answering $97$.