AMC 10 · 2019 · #20

Grade 7 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

parityprobability-basiccombinations-basicsystematic-enumerationpattern-recognition caseworksystematic-enumeration ↑ Prerequisites: parityprobability-basic

📏 Medium solution 💡 3 insights

Problem

Place the numbers 1, 2, …, 9 randomly into the nine cells of a 3 × 3 grid (one number per cell, each used once). What is the probability that the sum of the three numbers in every row and every column is odd?

Pick an answer.

(A)

$\frac{1}{21}$

(B)

$\frac{1}{14}$

(C)

$\frac{5}{63}$

(D)

$\frac{2}{21}$

(E)

$\frac{1}{7}$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): only parity matters — strip the values to O (odd) and E (even). Sub-question 1: which 3 × 3 O/E patterns make every row and column have odd sum? Sub-question 2: how many actual number-placements correspond to each valid pattern? Tool #2 (List): enumerate the valid parity patterns. Tool #16 (Complement-style insight): since 5 odds + 4 evens are fixed totals, the parity pattern is heavily constrained; the four evens must form a 2 × 2 rectangle.

1STEP 1

An odd row sum needs an odd count of odds, so the 5 odds split across 3 rows as 3 + 1 + 1: one all-odd row, two rows with a single odd.

5 = 3 + 1 + 1

2STEP 2

By the same column logic, the 4 evens land in the 2 × 2 block where the two non-special rows meet the two non-special columns.

evens form a 2 × 2 sub-grid

3STEP 3

Pick the all-odd row (3 ways) and the all-odd column (3 ways); this fixes the parity skeleton, giving 3 · 3 = 9 valid patterns.

#patterns = 3 · 3 = 9

4STEP 4

Fill the 5 odd cells with the odds (5!) and the 4 even cells with the evens (4!): 5! · 4! = 2880 arrangements per pattern.

#per pattern = 5! · 4! = 120 · 24 = 2880

5STEP 5

Favorable ÷ total: 9 · 5! · 4! = 25920 over 9! = 362880, which reduces to $\frac{1}{14}$ .

P =

\frac{9 · 5! · 4!}{9!}

\frac{25920}{362880}

\frac{1}{14}

→ (B)

Answer

\frac{1}{14}

Cross-check by simplifying differently:

\frac{9 · 5! · 4!}{9!}

\frac{9}{C(9, 4)}

\frac{9}{126}

\frac{1}{14}

. Here C(9, 4) = 126 counts the ways to choose which 4 of the 9 cells receive evens — only 9 of those (the 2 × 2 rectangles, formed by choosing 2 rows and 2 columns) are favorable. Sanity: choices like

\frac{1}{21}

\frac{5}{63}

would correspond to mis-counting the row-column independence;

\frac{1}{14}

is the unique result that matches the row-and-column 5 = 3 + 1 + 1 argument.

💡Key takeaway

This AMC 10 problem only needs Grade 7 parity-and-counting you already know — once you see the 4 evens must form a 2 × 2 rectangle, the count is C(3, 2)C(3, 2) = 9 patterns and P = $\frac{9}{C(9, 4)}$ = $\frac{1}{14}$ . The answer is (B).

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