Competition · AMC preparation · step 4 of 4

AMC 8 · 2008 · #14

Grade 7 counting
Archetype Casework by Position / Vertex Type └ Sub-archetype Arrangement Block Type
systematic-enumerationcombinations-basiccasework caseworksystematic-enumeration ↑ Prerequisites: systematic-enumerationmulti-digit-arithmetic
📏 Short solution 💡 3 insights 📊 Diagram
Problem
Three As, three Bs, and three Cs are placed in a 3 × 3 grid so that every row and every column contains exactly one of each letter. An A is already placed in the upper-left corner. How many different ways can the rest of the grid be filled?

Pick an answer.

(A)
2
(B)
3
(C)
4
(D)
5
(E)
6

AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Count Carefully

This is a counting problem with two grid rules, so Tool #13 (Count Carefully) drives the work. To avoid mistakes we use Tool #7 (Break into Subproblems): decide where the rest of the As go first, then the Bs, then the Cs. Tool #1 (Draw a Picture) keeps the 3 × 3 grid in front of us so we can see what is fixed and what is free at each step. The multiplication principle then turns the three subproblem counts into one product.

1STEP 1

Place the remaining As

The corner A fills row 1 and column 1, so the other two As sit in the bottom-right 2 × 2 block as a diagonal or anti-diagonal pair — 2 ways.

ways to place the remaining As = 2
2STEP 2

Place the Bs

Fix the A pattern; the top row still needs a B in one of two cells, and that single choice forces every other B — 2 ways.

ways to place the Bs (given the As) = 2
3STEP 3

Place the Cs

With an A and a B already in every row and column, the three empty cells each take the only letter left — 1 way for the Cs.

ways to place the Cs (given the As and Bs) = 1
4STEP 4

Multiply the three counts

Multiply the independent counts with the multiplication principle: 2 × 2 × 1 gives 4 grids in all.

total arrangements = 2 × 2 × 1 = 4 → (C)
Answer
4
List the four valid grids explicitly. With As on the main diagonal (cells (1,1),(2,2),(3,3)) the two grids are rows (A,B,C),(C,A,B),(B,C,A) and rows (A,C,B),(B,A,C),(C,B,A). With As on the anti-diagonal of the bottom 2 × 2 block (cells (1,1),(2,3),(3,2)) the two grids are rows (A,B,C),(B,C,A),(C,A,B) and rows (A,C,B),(C,B,A),(B,A,C). That is exactly 4 grids, all distinct. Cross-check: the total number of 3 × 3 Latin squares on three symbols is 12, and by symmetry exactly 123\frac{12}{3} = 4 of them have an A in the upper-left. Both checks agree.
💡Key takeaway

Place the letters one type at a time: 2 ways for the rest of the As, 2 ways for the Bs, then the Cs have no choice — 2 × 2 × 1 = 4 grids.

  • Place the remaining As
  • Place the Bs
  • Place the Cs
  • Multiply the three counts

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