AMC 10 · 2002 · #9
Grade 7 arithmeticUsing the letters A, M, O, S, and U, we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" USAMO occupies position
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The five letters $A$, $M$, $O$, $S$, $U$ are used exactly once each to build every possible five-letter arrangement. All of these arrangements are listed in dictionary (alphabetical) order. Find which numbered spot in that list the arrangement $USAMO$ lands in.
Givens: The letters available are $A$, $M$, $O$, $S$, $U$, each used once per word; Every one of the $5!=120$ arrangements is a valid "word"; The words are sorted in alphabetical order; Answer choices: (A) $112$, (B) $113$, (C) $114$, (D) $115$, (E) $116$
Unknowns: The position number of the word $USAMO$ in the sorted list
Understand
Restated: The five letters $A$, $M$, $O$, $S$, $U$ are used exactly once each to build every possible five-letter arrangement. All of these arrangements are listed in dictionary (alphabetical) order. Find which numbered spot in that list the arrangement $USAMO$ lands in.
Givens: The letters available are $A$, $M$, $O$, $S$, $U$, each used once per word; Every one of the $5!=120$ arrangements is a valid "word"; The words are sorted in alphabetical order; Answer choices: (A) $112$, (B) $113$, (C) $114$, (D) $115$, (E) $116$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems
Writing out all $120$ words would work but invites mistakes, so instead of hunting for $USAMO$ directly (Tool #16, Change Focus) we count how many words sit before it and add $1$. Tool #2 (Make a Systematic List) organizes those earlier words into clean alphabetical blocks: first by their opening letter, then by their second letter. Tool #7 (Identify Subproblems) handles the block sizes one position at a time — each time a letter is locked in, the leftover letters can be arranged in a factorial number of ways, and those counts simply add up. The one thing to guard is the final $+1$: the count of words before $USAMO$ is $114$, which is the decoy answer (C); the position itself is one more.
Execute — Answer: D
7.SP.C.8 Step 1 Count words that start before U
- Sort the letters: $A<M<O<S<U$.
- Any word whose first letter is $A$, $M$, $O$, or $S$ comes before every word starting with $U$, so all of those land before $USAMO$.
- For each of those $4$ opening letters, the remaining $4$ letters can be arranged in $4! = 24$ ways.
- That gives $4\times 24 = 96$ words before we even reach the $U$ block.
💡 Locking the first letter to something smaller than $U$ guarantees the word comes earlier, and the leftover letters fan out into $4!$ orderings.
7.SP.C.8 Step 2 Inside the U block, count by second letter
- Now look only at words starting with $U$.
- The target is $U\,S\,A\,M\,O$, so its second letter is $S$.
- The letters still available after $U$ are $\{A, M, O, S\}$, and in alphabetical order $A$, $M$, $O$ all come before $S$.
- Choosing any of those three as the second letter fixes the first two letters and leaves $3$ letters to arrange in $3! = 6$ ways.
- That is $3\times 6 = 18$ more words before $USAMO$.
💡 Within the $U$ block, a smaller second letter still puts the word earlier, so count those and multiply by the arrangements of what's left.
7.SP.C.8 Step 3 Check the remaining letters add nothing
- Fix the start as $U\,S$.
- The letters left are $\{A, M, O\}$ and the target continues $A$, then $M$, then $O$.
- At the third spot $A$ is already the smallest available, so no word beats it there; at the fourth spot $M$ is the smaller of $\{M, O\}$, again nothing earlier; the last letter $O$ is forced.
- So beyond the $18$ already counted, no further words come before $USAMO$.
💡 Once every remaining letter you pick is the smallest one left, there is nothing alphabetically earlier to count.
4.OA.A.3 Step 4 Add the counts and step up by one
- The total number of words before $USAMO$ is $96 + 18 = 114$.
- The position of $USAMO$ itself is one more than the number of words ahead of it, so it sits at $114 + 1 = 115$.
- That is choice (D).
- Watch the trap: $114$ is the count of words that come before, not the position — grabbing it directly gives the decoy (C).
💡 If $114$ words are ahead of you in line, you are standing in spot $115$.
7.SP.C.8 Sort the letters: $A<M<O<S<U$. Any word whose first letter is $A$, $M$, $O$, or 7.SP.C.8 Now look only at words starting with $U$. The target is $U\,S\,A\,M\,O$, so its 7.SP.C.8 Fix the start as $U\,S$. The letters left are $\{A, M, O\}$ and the target conti 4.OA.A.3 The total number of words before $USAMO$ is $96 + 18 = 114$. The position of $US Review
Reasonableness: The answer must be one of $115$'s neighbors near the top of the $120$-word list, and $USAMO$ starts with the last letter $U$, so it should sit deep in the list — well past $96$. The count breaks down as $96$ (words before the $U$ block) $+ 18$ (earlier words inside the $U$ block) $+ 1$ (USAMO itself) $= 115$, comfortably below the maximum position $120$. Every piece is a whole number and the total lands squarely in the answer range $112$–$116$, so $115$ is reasonable.
Alternative: Count from the end instead. The very last word alphabetically is $U\,S\,O\,M\,A$ at position $120$. Words starting $U\,S$ occupy the final $3! = 6$ positions, i.e. positions $115$–$120$. Listing just those six in order — $USAMO, USAOM, USMAO, USMOA, USOAM, USOMA$ — puts $USAMO$ first among them, at position $115$, confirming (D).
CCSS standards used (min grade 7)
7.SP.C.8Represent sample spaces and count outcomes using organized lists, tables, and tree diagrams (Counting words in alphabetical blocks by fixing one letter at a time and multiplying by the factorial number of arrangements of the leftover letters ($4\times 4!$ and $3\times 3!$).)4.OA.A.3Solve multistep word problems using the four operations (Adding the block counts $96+18=114$ and stepping up by one to get the position $115$.)
⭐ To find a word's spot in a sorted list, count how many words come before it by locking letters left to right, then add one for the word itself.
⭐ To find a word's spot in a sorted list, count how many words come before it by locking letters left to right, then add one for the word itself.
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