AMC 10 · 2021 · #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

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

Case B (no center) groups by corner count: three corners give 4 sets, two corners with one edge-midpoint give 4 more.

three corners: 4 sets; two corners + edge-mid: 4 sets

3STEP 3

All four edge-midpoints are pairwise non-adjacent: three-edge-mid sets give 4, one-corner-two-edge give 4 — 22 independent sets total.

three edge-mids: 4 sets; one corner + two edge-mids: 4 sets

4STEP 4

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

5STEP 5

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

6STEP 6

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

Answer

~36

Sanity. A naive count gives

\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.

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