2019 AMC 12A Problem 16

Question

The numbers

1,

2,

\ldots,

9

are randomly placed into the

9

squares of a

3 \times 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?

\dfrac{1}{21}

\dfrac{1}{14}

\dfrac{5}{63}

\dfrac{2}{21}

\dfrac{1}{7}

Answer: B

Concepts:parity basic probability permutations

Difficulty rating: 1800

Small hint:

There are

5

odd and

4

even numbers; each row and column needs an odd count of odd numbers

Big hint:

The odd entries must fill one entire row and one entire column, forming a plus shape of

5

cells

Solution:

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.