Printable

2022 AMC 10A Problem 9

2022 AMC 10A Problem 9AMC 10 Step by Step

Name: Date: October 28, 2026

Arrangements with Restrictions · difficulty 2 of 5 · about 3 min · position P1-10

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?

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

Answer

Key insight

The wide bottom-middle region touches all four others. Paint it first, then go around the other four in order: 4 choices for the first, then 3 for each.

Solution

1. Name the regions. Bottom row, left to right: a small piece $L$ , a wide middle piece $M$ , and a right piece $R$ . Top row: a left piece $U$ and a right piece $V$ .
2. Which regions touch? Reading the picture:
- $M$ touches all four others: $L$ , $R$ , $U$ and $V$ .
- $L$ touches $U$ ; $U$ touches $V$ ; $V$ touches $R$ .
- $L$ does not touch $R$ or $V$ , and $U$ does not touch $R$ .
3. Paint the busiest region first. $M$ touches everything, so it limits the most. It gets any of the $5$ colors.
4. Go around the other four in the order $L, U, V, R$ . Each one touches $M$ and the region painted just before it, and those two already have different colors:
- $L$ : any color except $M$ 's. $4$ choices.
- $U$ : not $M$ 's color and not $L$ 's. $3$ choices.
- $V$ : not $M$ 's and not $U$ 's. $3$ choices.
- $R$ : not $M$ 's and not $V$ 's. $3$ choices.
5. Multiply. $5 \times 4 \times 3 \times 3 \times 3 = 540$ .

The answer is $\boxed{\textbf{(D)}\ 540}$ .

Why this works

When counting colorings, choose an order in which each new region has a known number of already-painted neighbors, all with different colors. Starting with the region that touches everything does this: every later region sees exactly two different colors (or one, for the first), so each count is a fixed number and you simply multiply.

The trap

Misreading which regions touch, or thinking all five colors must be different, which gives 5 × 4 × 3 × 2 × 1 = 120.

Concise solution

Name the regions: the bottom row has a small left piece $L$ , a wide middle piece $M$ , and a right piece $R$ ; the top row has a left piece $U$ and a right piece $V$ . Reading the figure, $M$ touches all four others; $L$ touches $U$ ; $U$ touches $V$ ; $V$ touches $R$ . The pairs $L, R$ and $L, V$ and $U, R$ do not touch.

Color $M$ first: $5$ choices. Then color the others in the order $L, U, V, R$ , so that each new region touches exactly two already-colored regions, $M$ and the previous one, which always have different colors:

$L$ : any color but $M$ 's, $4$ choices.
$U$ : not $M$ 's or $L$ 's color, $3$ choices.
$V$ : not $M$ 's or $U$ 's color, $3$ choices.
$R$ : not $M$ 's or $V$ 's color, $3$ choices.

Total: $5 \cdot 4 \cdot 3 \cdot 3 \cdot 3 = 540$ .

The answer is $\boxed{\textbf{(D)}\ 540}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.5)
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 3: Correcting for Overcounting (§3.2, §3.4)
Introduction to Counting & Probability, Chapter 4: Committees and Combinations

Techniques

Set up the equation/formula and compute; no special trick needed