2019 AMC 12A Problem 16

There are $5$ odd and $4$ even numbers. Each row and column must contain an odd number of odd entries. The only way to place $5$ odd entries with every row and column odd is to fill one complete row and one complete column (a plus shape of $3 + 3 - 1 = 5$ cells). There are $3 \cdot 3 = 9$ such patterns. Each pattern admits $5!$ placements of the odd numbers and $4!$ of the even numbers, so the probability is $$\dfrac{9 \cdot 5! \cdot 4!}{9!} = \dfrac{1}{14}.$$ Thus, the correct answer is B.