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The number of permutation matrices of order $n$, where permutation matrices are given by $$A=[a_{i,j}],~a_{i,j}=0~\mbox{or}~1,~\mbox{so that}~\sum_{i=1}^na_{i,j}=\sum_{j=1}^na_{i,j}=1,~\forall i,j=1,2,...,n$$ Can it be $n!$ ?

Riaz
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  • One permutation matrix is related to one permutation, and vice versa. So the number is exactly $n!$. – Groups May 05 '19 at 11:19

1 Answers1

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The matrix is formed of $n$ rows, and in these rows you can put any of the following $n$ vectors, (without placing the same row in 2 different places) $$(1,0, \cdots, 0),(0,1,0,\cdots,0), \cdots, (0,\cdots,0,1)$$

So the question is the same as asking, if you have $n$ persons, in how many ways these people can sit on $n$ seats?

So yes the answer is $n!$

Fareed Abi Farraj
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