Competition · AMC preparation · step 4 of 4

AMC 8 · 2022 · #23

Grade 7 counting

Archetype Casework by Position / Vertex Type └ Sub-archetype Arrangement Block Type

combinations-basicsystematic-enumerationspatial-visualization caseworkcomplementary-counting ↑ Prerequisites: combinations-basicsystematic-enumeration

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Each of the 9 cells of a 3 × 3 grid is filled with either a triangle (△) or a circle (bigcirc). A "line" is any of the 3 rows, 3 columns, or 2 main diagonals — 8 lines total. Count how many of the 2⁹ = 512 possible fillings simultaneously contain a line of three △s AND a line of three bigcircs.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The headline question ("how many grids?") is too big to attack directly, but it cracks open the moment we use Tool #7 (Identify Subproblems) to ask a geometry sub-question first: which pairs of lines can be "all △" and "all ○" at the same time? Because the two lines cannot share a cell, only disjoint line pairs survive — and the only disjoint pairs in a 3 × 3 grid are two-different-rows or two-different-columns. Diagonals are eliminated immediately. That collapses the problem into a much smaller counting task. Tool #9 (Easier Related Problem) handles the remaining piece: count valid fillings under the column case alone, then double by symmetry. Tool #2 (Systematic List) supplies the casework on "how many monochrome columns appear" so nothing is missed or double-counted.

1STEP 1

Find which line pairs work

The △-line and bigcirc-line can't share a cell, so the only disjoint options are two different rows or two different columns.

disjoint pairs ∈ {row, row} ∪ {col, col}

2STEP 2

Use row-column symmetry

By row–column symmetry the row count equals the column count, and the two cases never overlap, so count the column case and multiply by 2.

Total = 2 × (configurations with a △-column AND a ○-column)

3STEP 3

Count two monochrome columns

Case 1 — one all-△ column and one all-bigcirc column: 3 × 2 ways to place them, times 6 mixed fillings of the last column, gives 36.

3 × 2 × (2³ - 2) = 3 × 2 × 6 = 36

Multiplying independent choices for each column is exactly the Grade 7 "compound events via organized lists" counting principle.

When exactly one column is all △ and exactly one other column is all ○, the number of valid grids is 36.

▸ Why?

Choosing the two solid columns and filling the leftover column are done independently, so the total is the number of ways to pick the solid columns times the number of ways to fill the leftover column.

▸ Why?

For each way to pick the two solid columns the very same list of leftover-column fillings is available, so the outcomes form that many equal groups and the count is groups times group-size.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

There are 6 ways to choose the solid columns: 3 choices for which column is all △, and for each of those, 2 columns remain to be the all ○ column, giving 3 × 2 = 6.

▸ Why?

Three choices, each followed by the same two further choices, make three equal groups of two, which total 3 × 2.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

There are 6 ways to fill the leftover column with a mix: its 3 cells give 2 × 2 × 2 = 8 possible fillings, and removing the 2 all-one-shape fillings leaves 6.

▸ Why?

Each of the 3 cells doubles the number of fillings, since every filling so far splits into an equal pair, one for each shape placed in the next cell, so the fillings build up as equal groups to 2 × 2 × 2 = 8.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

The 8 fillings divide with no overlap into the 2 that are all one shape and the rest that are mixed, so the mixed count is the whole 8 minus those 2 parts.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

7.SP.C.8Make A Systematic List

4STEP 4

Count three monochrome columns

Case 2 — all three columns monochrome with both shapes: 3 + 3 = 6. Adding Case 1 gives the column total 42.

3 + 3 = 6; 36 + 6 = 42 (column case)

5STEP 5

Double for the row case

Symmetry gives the row case another 42, and it's disjoint from the column case, so the total is 42 + 42 = 84 — choice (D).

42 + 42 = 84 → (D)

Answer

A quick sanity check on the size of the answer: the total number of 3 × 3 fillings is 2⁹ = 512. The number of grids with at least one monochrome △-line is roughly the same as the number with at least one monochrome bigcirc-line, and most of those do not coincide. Getting 84 valid grids (about 16% of 512) is in the right ballpark — definitely below 96 and well above 39. The cleaner internal check: the column case alone gives 36 + 6 = 42, which equals choice (B); doubling for the row case gives 84, which equals choice (D). The answer choices are deliberately spaced so that someone who forgets to double, or forgets the disjoint diagonal check, lands on 42 or 78 — both wrong.

💡Key takeaway

This AMC 8 problem only needs Grade 7 organized-list counting — multiplication and addition of cases — that you already know!

Find which line pairs work
Use row-column symmetry
Count two monochrome columns
Count three monochrome columns
Double for the row case

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Find which line pairs work

Use row-column symmetry

Count two monochrome columns

Count three monochrome columns

Double for the row case

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