AMC 10 · 2023 · #20

Grade 7 geometry-2d

combinations-basicsystematic-enumerationpermutations-basic identify-subproblemscaseworksystematic-enumeration ↑ Prerequisites: combinations-basicsystematic-enumeration

📏 Long solution 💡 3 insights 📊 Diagram

Problem

Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?

$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Each cell of a $3 \times 3$ grid is colored with one of four colors (red, white, blue, green) so that every $2 \times 2$ sub-grid contains all four colors. How many such colorings are there?

Givens: $3 \times 3$ grid (9 cells); Four colors: R, W, B, G; Every $2 \times 2$ sub-grid contains one cell of each color; A $3 \times 3$ grid contains four $2 \times 2$ sub-grids (top-left, top-right, bottom-left, bottom-right); Answer choices: (A) $24$, (B) $48$, (C) $60$, (D) $72$, (E) $96$

Unknowns: Total number of valid $3 \times 3$ colorings

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Make a Systematic List, #1 Draw a Diagram

The puzzle has a natural two-stage structure (Tool #7 — Identify Subproblems): (i) color the top-left $2 \times 2$ block in $4! = 24$ ways, then (ii) count how many ways the third column and third row can be filled so that all four $2 \times 2$ sub-grids contain all four colors. By symmetry across the $24$ initial choices, fix one (say R-W-B-G clockwise) and multiply at the end. For the completion step, the right column and bottom row each have $2$ "local" choices, giving $4$ combinations to check (Tool #2 — Systematic List). Tool #1 (Diagram) keeps the grid visible so the dependencies are easy to track.

Execute — Answer: D

#1 Draw a Diagram 3.G.A.2 Step 1

Sketch the $3 \times 3$ grid with cells $C_{11}, \dots, C_{33}$ and identify the four $2 \times 2$ blocks: top-left $\{C_{11},C_{12},C_{21},C_{22}\}$, top-right $\{C_{12},C_{13},C_{22},C_{23}\}$, bottom-left $\{C_{21},C_{22},C_{31},C_{32}\}$, bottom-right $\{C_{22},C_{23},C_{32},C_{33}\}$.
The center cell $C_{22}$ sits inside all four blocks.

$4$ overlapping $2 \times 2$ blocks, sharing the center $C_{22}$

💡 Drawing the four $2 \times 2$ windows on a $3 \times 3$ grid makes the overlap pattern (and the strong constraints) visible.

#7 Identify Subproblems 7.SP.C.8 Step 2

Stage 1: count colorings of the top-left $2 \times 2$ block.
Four distinct colors in four distinct cells gives $4! = 24$ ways.
Fix one representative arrangement to count completions, then multiply by $24$ at the end.

$$\#\{C_{11}, C_{12}, C_{21}, C_{22}\} = 4! = 24$$

💡 Every $2 \times 2$ block uses all four colors exactly once, so the top-left block is a permutation of $\{R, W, B, G\}$.

#1 Draw a Diagram 3.G.A.2 Step 3

Fix the representative: $C_{11} = R,\ C_{12} = W,\ C_{21} = B,\ C_{22} = G$.
Now the right column $(C_{13}, C_{23})$ and the bottom row $(C_{31}, C_{32})$ are each constrained by a $2 \times 2$ block that already has two colors fixed.

$$\begin{array}{|c|c|c|} \hline R & W & ? \\ \hline B & G & ? \\ \hline ? & ? & ? \\ \hline \end{array}$$

💡 Pinning down the top-left block reduces the remaining work to filling $5$ cells under heavy local constraints.

#7 Identify Subproblems 7.SP.C.8 Step 4

Top-right block contains $C_{12} = W$ and $C_{22} = G$, so the missing two colors $\{R, B\}$ fill $(C_{13}, C_{23})$ in one of two orders.
Bottom-left block contains $C_{21} = B$ and $C_{22} = G$, so the missing $\{R, W\}$ fill $(C_{31}, C_{32})$ in one of two orders.
That gives $2 \times 2 = 4$ local options — but they must also pass the bottom-right $2 \times 2$ check.

$$(C_{13}, C_{23}) \in \{(R,B), (B,R)\},\ (C_{31}, C_{32}) \in \{(R,W), (W,R)\}$$

💡 Each adjacent $2 \times 2$ block needs the two colors missing from its known half — naturally two choices for each missing pair.

#2 Make a Systematic List 4.OA.C.5 Step 5

Apply Tool #2 (Systematic List) — enumerate all $4$ combinations of the two binary choices and check the bottom-right block $\{C_{22}, C_{23}, C_{32}, C_{33}\}$ for a feasible $C_{33}$.

$$4 \text{ combinations} \to \text{count survivors}$$

💡 With only four cases, enumeration is faster than any clever argument — list, check, count.

#2 Make a Systematic List 7.SP.C.8 Step 6

Case 1: $(C_{13}, C_{23}) = (R, B)$, $(C_{31}, C_{32}) = (R, W)$.
Bottom-right block has $C_{22}=G, C_{23}=B, C_{32}=W$, three distinct — $C_{33}$ must be $R$.
✓ Case 2: $(R, B), (W, R)$.
Block has $G, B, R$ — distinct — $C_{33} = W$.
✓ Case 3: $(B, R), (R, W)$.
Block has $G, R, W$ — distinct — $C_{33} = B$.
✓ Case 4: $(B, R), (W, R)$.
Block has $C_{22}=G, C_{23}=R, C_{32}=R$ — two $R$s already, no valid $C_{33}$.
✗

$$3 \text{ valid completions out of } 4$$

💡 Three of the four binary-choice combos pass the bottom-right $2 \times 2$ check; the fourth forces a color collision.

#7 Identify Subproblems 4.OA.A.2 Step 7

Multiply across stages: $24$ top-left colorings $\times\ 3$ valid completions per top-left coloring.

$$24 \times 3 = 72 \;\Rightarrow\; \textbf{(D) }72$$

💡 Multiplication principle: independent stages multiply, and the $3$-completion count is the same for every top-left coloring by symmetry.

[1] #1 3.G.A.2 Sketch the $3 \times 3$ grid with cells $C_{11}, \dots, C_{33}$ and identify the

[2] #7 7.SP.C.8 Stage 1: count colorings of the top-left $2 \times 2$ block. Four distinct color

[3] #1 3.G.A.2 Fix the representative: $C_{11} = R,\ C_{12} = W,\ C_{21} = B,\ C_{22} = G$. Now

[4] #7 7.SP.C.8 Top-right block contains $C_{12} = W$ and $C_{22} = G$, so the missing two color

[5] #2 4.OA.C.5 Apply Tool #2 (Systematic List) — enumerate all $4$ combinations of the two bina

[6] #2 7.SP.C.8 Case 1: $(C_{13}, C_{23}) = (R, B)$, $(C_{31}, C_{32}) = (R, W)$. Bottom-right b

[7] #7 4.OA.A.2 Multiply across stages: $24$ top-left colorings $\times\ 3$ valid completions pe

Review

Reasonableness: Spot-check: a row-shift coloring like $\begin{smallmatrix} R & W & R \\ B & G & B \\ R & W & R \end{smallmatrix}$ has top-left, top-right, bottom-left, bottom-right blocks all reading $\{R, W, B, G\}$ — valid. The constraint forces every row to be a $2$-coloring that repeats with period $2$ (and same for every column), so the top-left $2 \times 2$ pins down the entire pattern *up to* a choice of which two colors form the alternate row/column — that secondary choice is exactly the $3$ completions counted above. $24 \times 3 = 72$ matches answer (D) and is comfortably between (B) $48$ and (E) $96$.

Alternative: Tool #9 (Solve an Easier Related Problem): try a $2 \times 3$ grid first. Top-left $2 \times 2$: $4! = 24$ ways. The third column $(C_{13}, C_{23})$ must contain the two colors missing from the top-right $2 \times 2$ — exactly $2$ choices. Total $24 \times 2 = 48$. For $3 \times 3$, do the third row independently: $2$ choices for $(C_{31}, C_{32})$, but only $3$ of the $4$ combined options keep the bottom-right block valid. So $24 \times 3 = 72$ — same (D).

CCSS standards used (min grade 7)

3.G.A.2 Partition shapes into parts with equal areas and express the area of each part as a unit fraction of the whole (Decomposing the $3 \times 3$ grid into its four overlapping $2 \times 2$ sub-grids and tracking which cells each one contains.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting permutations of four colors in a $2 \times 2$ block ($4! = 24$) and combining the independent choice counts via the multiplication principle.)
4.OA.C.5 Generate a number or shape pattern that follows a given rule (Enumerating the $4$ binary-choice combinations for $(C_{13}, C_{23})$ and $(C_{31}, C_{32})$ in a fixed order.)
4.OA.A.2 Multiply or divide to solve word problems involving multiplicative comparison (Combining $24$ top-left colorings with $3$ valid completions each via multiplication: $24 \times 3 = 72$.)

⭐ The top-left $2 \times 2$ block can be colored in $4! = 24$ ways. For each choice, the right column and bottom row each have $2$ binary options, but only $3$ of the $4$ combined options keep the bottom-right $2 \times 2$ valid. Multiply: $24 \times 3 = \textbf{(D) }72$.