AMC 10 · 2019 · #20
Grade 7 geometry-2dProblem
The numbers 1,2,…,9 are randomly placed into the 9 squares of a 3×3 grid. Each square gets one number, and each of the numbers is used once. What is the probability that the sum of the numbers in each row and each column is odd?
Pick an answer.
AMC 10 2019 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.
Toolkit + CCSS Solution
Understand
Restated: Place the numbers $1, 2, \ldots, 9$ randomly into the nine cells of a $3 \times 3$ 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?
Givens: Numbers $1$ through $9$: five odd ($1, 3, 5, 7, 9$) and four even ($2, 4, 6, 8$); All $9!$ arrangements are equally likely; Win condition: every row sum is odd AND every column sum is odd; Choices: (A) $\tfrac{1}{21}$, (B) $\tfrac{1}{14}$, (C) $\tfrac{5}{63}$, (D) $\tfrac{2}{21}$, (E) $\tfrac{1}{7}$
Unknowns: Probability that all six row/column sums are simultaneously odd
Understand
Restated: Place the numbers $1, 2, \ldots, 9$ randomly into the nine cells of a $3 \times 3$ 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?
Givens: Numbers $1$ through $9$: five odd ($1, 3, 5, 7, 9$) and four even ($2, 4, 6, 8$); All $9!$ arrangements are equally likely; Win condition: every row sum is odd AND every column sum is odd; Choices: (A) $\tfrac{1}{21}$, (B) $\tfrac{1}{14}$, (C) $\tfrac{5}{63}$, (D) $\tfrac{2}{21}$, (E) $\tfrac{1}{7}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #16 Change Focus / Count the Complement
Tool #7 (Subproblems): only parity matters — strip the values to O (odd) and E (even). Sub-question 1: which $3 \times 3$ 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 $2 \times 2$ rectangle.
Execute — Answer: B
2.OA.C.3 Step 1 - Each row sums to an odd number iff it contains an odd count of odds — that is, either $1$ odd or $3$ odds.
- Since there are only $5$ odds spread across $3$ rows, the row-odd-counts (each in $\{1, 3\}$) summing to $5$ force the split $5 = 3 + 1 + 1$ (in some order).
- So exactly one row is all-odd, and the other two rows have exactly one odd each (and therefore two evens each).
💡 An odd row sum needs an odd number of odd entries — and $5$ odds split as $3+1+1$.
2.OA.C.3 Step 2 - The same logic applies column-wise: exactly one column is all-odd, the other two columns have one odd and two evens each.
- Combining: the $4$ evens lie in the intersection of the $2$ "non-odd" rows and the $2$ "non-odd" columns — that is, the evens occupy a $2 \times 2$ block (the four cells where neither the all-odd row nor the all-odd column meets).
💡 Two non-special rows $\times$ two non-special columns $=$ exactly $4$ cells for the $4$ evens.
7.SP.C.8 Step 3 - Count parity patterns: choose which row is all-odd ($3$ ways) and which column is all-odd ($3$ ways).
- That uniquely determines the parity skeleton.
- So there are $3 \cdot 3 = 9$ valid parity patterns.
💡 Independent choices: pick the all-odd row and the all-odd column.
7.SP.C.8 Step 4 - For each parity pattern, count the actual number assignments.
- Place the $5$ odd numbers into the $5$ odd-positions: $5!$ orderings.
- Place the $4$ even numbers into the $4$ even-positions: $4!$ orderings.
- So each pattern produces $5! \cdot 4!$ arrangements.
💡 Odd numbers fill odd cells in any order; same for evens — independent choices.
7.SP.C.8 Step 5 - Total favorable arrangements: $9 \cdot 5! \cdot 4! = 9 \cdot 2880 = 25920$.
- Total possible arrangements: $9! = 362880$.
- Probability $= \frac{25920}{362880} = \frac{1}{14}$.
💡 Favorable $\div$ total — and the factorials cancel cleanly.
2.OA.C.3 Each row sums to an odd number iff it contains an odd count of odds — that is, e 2.OA.C.3 The same logic applies column-wise: exactly one column is all-odd, the other two 7.SP.C.8 Count parity patterns: choose which row is all-odd ($3$ ways) and which column i 7.SP.C.8 For each parity pattern, count the actual number assignments. Place the $5$ odd 7.SP.C.8 Total favorable arrangements: $9 \cdot 5! \cdot 4! = 9 \cdot 2880 = 25920$. Tota Review
Reasonableness: Cross-check by simplifying differently: $\frac{9 \cdot 5! \cdot 4!}{9!} = \frac{9}{\binom{9}{4}} = \frac{9}{126} = \frac{1}{14}$. Here $\binom{9}{4} = 126$ counts the ways to choose which $4$ of the $9$ cells receive evens — only $9$ of those (the $2 \times 2$ rectangles, formed by choosing $2$ rows and $2$ columns) are favorable. Sanity: choices like $\frac{1}{21}$ or $\frac{5}{63}$ 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.
Alternative: Tool #16 (Change focus to the evens): the only choice that matters is which $4$ cells hold the evens (the odds fall into the rest). Out of $\binom{9}{4} = 126$ ways to pick those cells, the favorable ones are exactly the $2 \times 2$ axis-aligned rectangles — $\binom{3}{2}\binom{3}{2} = 9$ of them. So $P = \frac{9}{126} = \frac{1}{14}$.
CCSS standards used (min grade 7)
2.OA.C.3Determine whether a group of objects has an odd or even number (Recognizing that a row sum is odd iff the row contains $1$ or $3$ odd entries; splitting $5 = 3 + 1 + 1$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting favorable arrangements ($9 \cdot 5! \cdot 4!$) versus total ($9!$) to get the probability.)
⭐ This AMC 10 problem only needs Grade 7 parity-and-counting you already know — once you see the $4$ evens must form a $2 \times 2$ rectangle, the count is $\binom{3}{2}\binom{3}{2} = 9$ patterns and $P = \frac{9}{\binom{9}{4}} = \frac{1}{14}$. The answer is $\textbf{(B)}$.
⭐ This AMC 10 problem only needs Grade 7 parity-and-counting you already know — once you see the $4$ evens must form a $2 \times 2$ rectangle, the count is $\binom{3}{2}\binom{3}{2} = 9$ patterns and $P = \frac{9}{\binom{9}{4}} = \frac{1}{14}$. The answer is $\textbf{(B)}$.
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