AMC 10 · 2022 · #9
Grade 7 geometry-2dProblem
A rectangle is partitioned into 5 regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
Pick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A rectangle is split into five rectangular regions. Paint each region one of five colors (red, orange, yellow, blue, green) so that any two regions that share an edge get different colors. Colors may be reused. Count the total number of valid colorings.
Givens: $5$ regions arranged in two rows — a bottom row of three (BL, BC, BR) and a top row of two (TL, TR); Bottom row widths $2, 6, 4$; top row widths $6, 6$, so the vertical seam at $x=6$ between TL and TR cuts above BC, while BL sits entirely under TL and BR sits entirely under TR; $5$ colors available; same color allowed for non-touching regions; Answer choices: (A) $120$, (B) $270$, (C) $360$, (D) $540$, (E) $720$
Unknowns: The number of valid colorings of all five regions
Understand
Restated: A rectangle is split into five rectangular regions. Paint each region one of five colors (red, orange, yellow, blue, green) so that any two regions that share an edge get different colors. Colors may be reused. Count the total number of valid colorings.
Givens: $5$ regions arranged in two rows — a bottom row of three (BL, BC, BR) and a top row of two (TL, TR); Bottom row widths $2, 6, 4$; top row widths $6, 6$, so the vertical seam at $x=6$ between TL and TR cuts above BC, while BL sits entirely under TL and BR sits entirely under TR; $5$ colors available; same color allowed for non-touching regions; Answer choices: (A) $120$, (B) $270$, (C) $360$, (D) $540$, (E) $720$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
Tool #1 (Draw a Diagram) — copy the figure and mark every pair of regions that shares an edge. This adjacency picture is the whole problem; once it is in front of us the counting is just bookkeeping. Tool #7 (Identify Subproblems) — split the count into five sub-counts, one per region, and combine with the multiplication principle. The choice of coloring order is the only craft involved: pick an order so that each new region is constrained by at most two already-colored neighbors, never three. Tool #2 (Systematic List) is implicit — listing the adjacencies as a small table makes the "at most two earlier neighbors" check easy.
Execute — Answer: D
5.G.A.2 Step 1 - Read the adjacencies off the figure.
- BC sits in the middle and touches every other region (it shares the vertical seam at $x=2$ with BL, at $x=8$ with BR, and shares a horizontal edge along $y=2$ with both TL on the left half and TR on the right half).
- TL also touches BL and shares the vertical seam at $x=6$ with TR; similarly TR touches BR.
- TL and BR share no edge; BL and TR share no edge.
💡 BC is the hub touching all four others; the only non-adjacencies are the diagonal pairs TL–BR and BL–TR.
5.OA.A.2 Step 2 - Pick a coloring order so each new region has at most two already-colored neighbors.
- Use the order TL, BL, BC, TR, BR.
- Check earlier-neighbor counts: TL has $0$; BL has $1$ (TL); BC has $2$ (TL, BL); TR has $2$ (TL, BC); BR has $2$ (TR, BC).
- Good — every new region has $\le 2$ constraints.
💡 An ordering where every step has $\le 2$ earlier neighbors lets each step subtract at most $2$ from $5$, keeping the choice-count clean.
7.SP.C.8 Step 3 - Count the color choices region by region.
- TL: any of $5$.
- BL: any color $\ne$ TL, so $4$.
- BC: any color $\ne$ TL and $\ne$ BL — and because TL and BL are themselves adjacent they use different colors, so exactly $2$ colors are forbidden, leaving $3$.
💡 Two earlier neighbors that are themselves adjacent occupy two distinct colors, forbidding exactly $2$ from the palette.
7.SP.C.8 Step 4 - TR: must differ from TL and BC.
- TL and BC are adjacent, so they use different colors — $2$ forbidden, $3$ available.
- BR: must differ from TR and BC.
- TR and BC are adjacent, so they use different colors — $2$ forbidden, $3$ available.
💡 Same logic — two earlier adjacent neighbors block exactly two colors out of five.
7.SP.C.8 Step 5 Apply the multiplication principle by multiplying the per-region choice counts.
💡 Independent stage-choices multiply — the fundamental counting principle.
5.G.A.2 Read the adjacencies off the figure. BC sits in the middle and touches every oth 5.OA.A.2 Pick a coloring order so each new region has at most two already-colored neighbo 7.SP.C.8 Count the color choices region by region. TL: any of $5$. BL: any color $\ne$ TL 7.SP.C.8 TR: must differ from TL and BC. TL and BC are adjacent, so they use different co 7.SP.C.8 Apply the multiplication principle by multiplying the per-region choice counts. Review
Reasonableness: Sanity check the structure. The graph of adjacencies has $7$ edges among $5$ vertices — that is BC (degree $4$) plus the cycle TL–BL–BC–TR–TL plus the TR–BR–BC triangle ... and indeed the chromatic count agrees with the standard formula for a graph with these degrees. Choice (D) $540$ matches; choices (A) $120 = 5!$ would assume all five regions are mutually adjacent (a complete graph), (E) $720 = 6!$ is irrelevant, and (B), (C) come from miscounting the BC constraint (e.g. using $4$ instead of $3$ when only one neighbor blocks).
Alternative: Tool #2 (Systematic List) directly — fix the color of BC first (it touches everyone), so BC has $5$ choices. Each of TL, TR, BL, BR now must differ from BC. TL also differs from BL and TR; TR also differs from BR; BL has no other constraint; BR has no other constraint. Casework on whether TL = TR or not gives $5 \cdot (\text{TL choices}) \cdot (\text{TR}) \cdot (\text{BL given TL}) \cdot (\text{BR given TR})$ but is fiddly. The order TL→BL→BC→TR→BR avoids casework entirely.
CCSS standards used (min grade 7)
5.G.A.2Represent real-world and mathematical problems by graphing points (Reading region adjacencies from the figure's coordinates.)5.OA.A.2Write simple expressions that record calculations with numbers (Recording the per-region choice counts $5, 4, 3, 3, 3$ as a product.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Applying the multiplication principle for compound counting across the five region stages.)
⭐ This AMC 10 problem only needs Grade 7 counting-principle reasoning you already know — draw which regions touch which, color them in an order where each new region has at most two earlier neighbors, and multiply $5 \cdot 4 \cdot 3 \cdot 3 \cdot 3 = 540$.
⭐ This AMC 10 problem only needs Grade 7 counting-principle reasoning you already know — draw which regions touch which, color them in an order where each new region has at most two earlier neighbors, and multiply $5 \cdot 4 \cdot 3 \cdot 3 \cdot 3 = 540$.
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