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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 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.

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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: $24 \times 3 = \textbf{(D) }72$.