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We want to find the number of $n × n$ binary matrices that have precisely one 1 in every row and every column. I believe the number of such matrices is $n!$. This is my reasoning: the number of $n × n$ binary matrices that have precisely one 1 in every row and every column corresponds to all possible permutations of the identity matrix (i.e. every permutation matrix of size $n$), of which there are $n!$. How would one go about proving this?

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closed as off-topic by Namaste, Leucippus, Xander Henderson, Jendrik Stelzner, user91500 Aug 23 '18 at 9:13

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  • $\begingroup$ Do you ever hear about permutation? $\endgroup$ – Cloud JR Aug 22 '18 at 13:32
  • $\begingroup$ Yes. I'm assuming the number of such matrices would correspond to all possible permutations of the identity matrix, correct? $\endgroup$ – Frank Aiello Aug 22 '18 at 13:33
  • $\begingroup$ You are correct $\endgroup$ – Cloud JR Aug 22 '18 at 13:38
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The first row can be filled in $n$ ways ($n$ columns is possible to put $1$ and $0$ in other column ). After filling first row there is $(n-1)$ ways (One column is already used) as so on.

BY FUNDAMENTAL PRINCIPLE OF COUNTING, there are $n!$ ways.

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