2019 AMC 10A Problem 20

Only the parity of each entry matters. There are five odd numbers ($1, 3, 5, 7, 9$) and four even ones, and every choice of which $5$ cells hold the odd numbers is equally likely: $\binom95 = 126$ patterns.

A row sum is odd exactly when the row contains $1$ or $3$ odd numbers. The three row counts are odd and add to $5$, so they must be $3, 1, 1$ in some order: one row is entirely odd and each other row has exactly one odd entry. The all-odd row gives every column one odd number; the two remaining odd entries must keep each column count odd, so they cannot land in different columns (that would make two columns even). They therefore sit in the same column, making that column all odd.

So a good pattern is a full row plus a full column of odds, a "cross," and any cross uses exactly $3 + 3 - 1 = 5$ cells. There are $3 \cdot 3 = 9$ crosses, and each does work (the full row and column have sums of three odds; the other rows and columns have exactly one odd).

Probability: $\frac{9}{126} = \frac{1}{14}$.

The answer is $\boxed{\textbf{(B)}\ \frac{1}{14}}$.

Conditions about sums being odd or even depend only on parities, so replace the $9!$ arrangements by the $\binom95$ odd/even patterns, each equally likely. Then a counting argument (odd counts summing to $5$ in rows, then in columns) forces the shape. Reducing a permutation problem to a pattern problem is the standard first move for parity questions.

Count full arrangements: $9$ crosses, $5!$ ways to place the odd numbers on the cross, $4!$ ways for the evens, over $9!$ total: $\frac{9 \cdot 120 \cdot 24}{362880} = \frac{1}{14}$.

• Counting the odd-number patterns as 9 but dividing by 9! or by 5! x 4! instead of by the C(9,5) = 126 equally likely odd/even patterns.
• Allowing rows with two odd numbers, or forgetting that the column condition forces the two leftover odds into the same column.