My attempt is quite handwaivy. But I think this has something to do with permutation matrices. I am absolutely new to this topic. can anyone throw any light on this solution? I know there are $2^{n\times m}$ binary matrices of size $n\times m$, and $n!m!$ possible permutations, but somehow I fail to get an intuition on what this implies for the equivalence classes.
2026-03-25 11:04:53.1774436693
Find the number of $6 \times 7$ matrices with entries $[0,1]$ such that their row and column sums are odd.
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If the row sums are all odd, then the total number of $1$s is the sum of these $6$ odd numbers, hence even. If the column sums are all odd, then the total number of $1$s is the sum of these $7$ odd numbers, hence odd. A contradiction; hence the number of such matrices is $0$.