Find the number of $6 \times 7$ matrices with entries $[0,1]$ such that their row and column sums are odd. [closed]

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.

closed as off-topic by Saad, Shailesh, Xander Henderson, mrtaurho, InterstellarProbeApr 12 at 16:14

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• So the entry could be 1/2? – Maria Mazur Feb 18 at 10:28

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 $$0$$.
• @InterstellarProbe If the column sums are odd, then the sum of sums is odd because there are $7$ columns. – Servaes Apr 12 at 17:40