AMC 10 · 2022 · #9

Grade 7 geometry-2d
combinations-basiccaseworksystematic-enumeration caseworkidentify-subproblems ↑ Prerequisites: combinations-basic
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
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.

Pick an answer.

(A)
120
(B)
270
(C)
360
(D)
540
(E)
720

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

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.

1STEP 1

Read the adjacencies off the figure: BC touches all four other regions; the only non-touching pairs are the diagonals TL–BR and BL–TR.

Adjacencies: {TL - BL, TL - BC, TL - TR, TR - BC, TR - BR, BL - BC, BC - BR}
2STEP 2

Color in the order TL, BL, BC, TR, BR — then each new region has at most two already-colored neighbors (counts 0, 1, 2, 2, 2).

order: TL→BL→BC→TR→BR; earlier-neighbors: 0,1,2,2,2
3STEP 3

TL is free (5 choices); BL avoids TL (4); BC avoids both TL and BL, which differ, so exactly two are blocked, leaving 3.

TL:5, BL:4, BC:3
4STEP 4

TR avoids TL and BC (adjacent, so different colors) — 3 left; BR avoids TR and BC (adjacent) — 3 left.

TR:3, BR:3
5STEP 5

Multiply the per-region choice counts by the multiplication principle to reach the total of 540 colorings.

5 × 4 × 3 × 3 × 3 = 20 × 27 = 540 → (D)
Answer
540
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).
💡Key takeaway

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 · 4 · 3 · 3 · 3 = 540.