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, InterstellarProbe Apr 12 at 16:14

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  • 3
    $\begingroup$ So the entry could be 1/2? $\endgroup$ – 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 $0$.

  • $\begingroup$ @InterstellarProbe If the column sums are odd, then the sum of sums is odd because there are $7$ columns. $\endgroup$ – Servaes Apr 12 at 17:40
  • $\begingroup$ @InterstellarProbe To reach a contradiction. Have you read my answer? $\endgroup$ – Servaes Apr 12 at 19:25

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