AMC 10 · 2022 · #7
Grade 7 countinggeometry-2d
Pick an answer.
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.
List the adjacencies
Write down which regions touch.
BC is the hub touching all four others; the only non-adjacencies are the diagonal pairs TL–BR and BL–TR.
5.G.A.2Draw A DiagramChoose the painting order
A good order makes each step simple.
An ordering where every step has ≤ 2 earlier neighbors lets each step subtract at most 2 from 5, keeping the choice-count clean.
5.OA.A.2Identify SubproblemsPaint the first three
The first three have shrinking choices.
Two earlier neighbors that are themselves adjacent occupy two distinct colors, forbidding exactly 2 from the palette.
7.SP.C.8Identify SubproblemsPaint the last two
Each of the last two has three choices.
Same logic — two earlier adjacent neighbors block exactly two colors out of five.
7.SP.C.8Identify SubproblemsMultiply
Multiplying gives 540.
Independent stage-choices multiply — the fundamental counting principle.
Independent stage choices multiply, so the whole count is one product.
▸ Why?
Each stage is chosen freely given the earlier ones, so every combination is possible.
▸ Why?
Each finished colouring arises from exactly one run through the stages, so nothing is counted twice.
This AMC 12 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.