AMC 10 · 2002 · #9

Grade 7 arithmetic
Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count
permutations-basicfactorialsystematic-enumeration systematic-enumeration ↑ Prerequisites: permutations-basic
📏 Medium solution 💡 2 insights
Problem
The five letters A, M, O, S, U are used exactly once each to build every possible five-letter arrangement, and all of these arrangements are listed in dictionary (alphabetical) order. Find which numbered spot in that list the arrangement USAMO lands in.

Pick an answer.

(A)
112
(B)
113
(C)
114
(D)
115
(E)
116

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

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.

1STEP 1

Count words that start before U

Letters rank A ＜ M ＜ O ＜ S ＜ U, so every word opening with A, M, O, or S beats USAMO: 4 × 4! = 96 words.

4 × 4! = 4 × 24 = 96
2STEP 2

Inside the U block, count by second letter

Among words starting with U, second letters A, M, O all precede USAMO's S: 3 × 3! = 18 more words.

3 × 3! = 3 × 6 = 18
3STEP 3

Check the remaining letters add nothing

Fix the start as U S. Then A, M, O are each the smallest letter still left, so nothing extra slips in earlier.

third, fourth, fifth letters each smallest available → 0 extra words
4STEP 4

Add the counts and step up by one

96 + 18 = 114 words come first, so USAMO sits at 114 + 1 = 115, choice (D); 114 itself is the decoy.

96 + 18 = 114, 114 + 1 = 115 → (D)
Answer
115
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.
💡Key takeaway

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.

  • Count words that start before U
  • Inside the U block, count by second letter
  • Check the remaining letters add nothing
  • Add the counts and step up by one

More like this

Same archetype — closest grade level first.