Correct : c

The correct answer is Option C: 512.
The key idea is to figure out how many entries can be chosen freely, with the rest being determined by the parity constraints.
Consider filling the top-left 3×3 submatrix (the first 3 rows and first 3 columns) with any combination of 0s and 1s. There are 2⁹ = 512 ways to do this.
Once these 9 entries are fixed:
• The 4th entry in each of the first 3 rows is forced - it must make the row sum even.
• The entire 4th row is forced - each entry must make its column sum even.
• The bottom-right entry (row 4, column 4) is automatically consistent with both the row and column parity constraints, because the total sum of the matrix (mod 2) must be 0, which follows from all row sums being even.
So every free choice of the 3×3 submatrix gives exactly one valid 4×4 matrix, and every valid matrix corresponds to exactly one such choice.
Total = 2⁹ = 512.