Problem 14 · 2008 AMC 8
Medium
Counting & Probability
latin-square
Three A's, three B's, and three C's are placed in the nine spaces so that each row and column contain one of each letter. If A is placed in the upper left corner, how many arrangements are possible?
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Answer: C — 4 arrangements.
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Hint 1 of 2
"One of each per row and column" means once you commit a couple of cells, the rest are forced — so just count the genuinely free choices.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the multiplication principle: multiply the number of choices at each free decision point.
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Approach: count only the free choices, let the rest cascade
- A sits top-left. The rest of row 1 is B, C in some order — 2 ways. Now look at the cell below A (start of row 2): it can't be A, so it's B or C — 2 ways.
- After those two picks, every remaining cell is forced (each row and column still needs its missing letter). So the count is just 2 × 2 = 4.
- Why this transfers: in tight grid/arrangement puzzles, find the few cells you freely choose and multiply — the constraints fill in the rest for free, so you never enumerate all the boards.
Another way — list them:
- Fixing row 1 as A B C, the two valid completions are (B C A / C A B) and (C A B / B C A). Fixing row 1 as A C B gives two more.
- That's 2 + 2 = 4 arrangements total.
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