Competition · AMC preparation · step 4 of 4

AMC 10 · 2021A · #25

Grade 5 counting
Archetype Casework by Position / Vertex Type └ Sub-archetype Grid / Solid Cell Position
systematic-enumerationcombinations-basiccaseworksymmetry-argument caseworksystematic-enumeration ↑ Prerequisites: systematic-enumeration
📏 Long solution 💡 3 insights
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Problem
Place 3 indistinguishable red chips, 3 indistinguishable blue chips, and 3 indistinguishable green chips into the 9 squares of a 3 × 3 grid (one chip per square) so that no two chips of the same color share an edge (horizontally or vertically adjacent). How many such placements are there?

Pick an answer.

(A)
12
(B)
18
(C)
24
(D)
30
(E)
36

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

Tool #9 (Easier Problem) — first ask the easier subquestion: how many ways can ONE color (say Red) be placed so that no two reds are adjacent? Tool #2 (Systematic List) — list all 3-cell independent sets of the 3 × 3 grid; this turns out to be a small finite set. Tool #1 (Diagram) — sketch each independent set so the patterns are visible. Tool #7 (Subproblems) — for each Red layout, count the ways to color the remaining 6 cells with 3 B + 3 G (a much easier subproblem). Tool #3 (Eliminate) — the answer choices {12, 18, 24, 30, 36} all factor through small multiples, helping us spot the structure.

1STEP 1

Count case A

Split on the center: a color using center (2,2) needs its other two chips in corners, so C(4, 2) = 6 Case A sets.

Case A (contains center): C(4, 2) = 6 sets
2STEP 2

Count case B

Case B (no center) gives four families of 4 — the edge-midpoints are pairwise non-adjacent — for 22 independent sets.

three corners: 4 sets; two corners + edge-mid: 4 sets three edge-mids: 4 sets; one corner + two edge-mids: 4 sets
3STEP 3

Keep the two workable patterns

Only 6 of the 22 triples extend to a full 3-coloring: 2 diagonal shapes (P1) and 4 center-plus-corner-pair shapes (P2).

P1 (diagonals): 2 shapes; P2 (center + corner-pair): 4 shapes
4STEP 4

Count fillings for P1

Fix Red on a diagonal (P1): every B/G choice cascades from one, leaving 2 fillings, so 2 · 3 · 2 = 12 colorings.

P1 count = 2 · 3 · 2 = 12
5STEP 5

Count fillings for P2

P2 works the same way — the Red layout forces the fill up to a B⇔G swap — giving 4 · 3 · 2 = 24 colorings.

P2 count = 4 · 3 · 2 = 24
6STEP 6

Add the two patterns

P1 (Red on a diagonal) and P2 (center plus two corners) cannot both occur, so the counts add: 12 + 24 = 36 → (E).

Total = 12 + 24 = 36 → (E)
Answer
36
Sanity. A naive count gives 9!3!\frac{9!}{3!} 3! 3! = 1680 unrestricted colorings; the answer 36 is roughly 2%, which feels right for a tight no-adjacency constraint. Symmetry check: the grid has D₄ symmetry (rotations + reflections), giving 8 symmetries. The diagonal pattern set has 2 orbits of 1 each under D₄ swap (2 diagonals are exchanged by reflection); the corner-pair pattern has 4 orbits (top/bottom/left/right merge under D₄). With 3 color labels, 12 + 24 = 36 is consistent with the symmetry counts. The B⇔G swap doubling (within fixed Red layout) appears twice and combines correctly.
💡Key takeaway

This AMC 10 problem only needs Grade 5 systematic listing and multiplication you already know — try all valid placements of one color (Red): only 2 diagonals and 4 "center + same-side-corner-pair" shapes work; for each, exactly 2 ways to fill the other six cells with B and G; multiply by 3 color choices and add — 2 · 3 · 2 + 4 · 3 · 2 = 12 + 24 = 36.

  • List the independent sets
  • Recount case B
  • Finish case B
  • Total the independent sets
  • Count fillings for P1
  • Count fillings for P2

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