Probability of a grid with row and column

Question

Put the nine different numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ at random in nine different places in a $3×3$ grid. What is the probability that each row sum and column sum is odd?

Answer 1

Hint.

1. Prove that one row has to be full of odd numbers, and one column has to be full of odd numbers, and except that row and that column all numbers are even.
2. Compute the number of ways to choose which row and which column are going to be the odd ones.
3. Compute the number of ways to fill the chosen row/column with the odd numbers and the remaining placeholders with even numbers.
4. Do the relevant multiplications and divide by the total number of ways to fill the grid arbitrarily.

Answer 2

Hint: Exactly one row must be all odds and exactly one column must be all odds. That leaves the remaining $4$ slots to be even numbers. How many combinations are there?

Hint 2: See Mostafa Ayaz's diagram. How many ways can you place the two orange lines? How many ways can you arrange the numbers within those orange lines? How many ways can you arrange the remaining four blues.?

Answer 3

There are 4 even numbers and 5 odd numbers so the only valid cases are

each containing $5!\times 4!$ different cases. So we have $5\times 5!\times 4!=14400$ different cases in total.