Consider a square matrix of order 5 such that | vedclass

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(A) square matrix of order $n=5$ where each row and each column has exactly one non-zero element (which is $1$) is a permutation matrix. There are $n!

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$44$

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(A) square matrix of order $n=5$ where each row and each column has exactly one non-zero element (which is $1$) is a permutation matrix. There are $n!$ such matrices in total. For $n=5$,the total number of such matrices is $5! = 120$. However,we are given the condition $a_{ij} = 0$ whenever $i + j = n + 1 = 6$. This means the non-zero element cannot be placed on the anti-diagonal of the matrix. Let $S_n$ be the set of all permutation matrices of order $n$. We want to find the number of matrices in $S_n$ such that no $1$ is on the anti-diagonal. This is equivalent to finding the number of permutations $\sigma$ of $\{1, 2, 3, 4, 5\}$ such that $\sigma(i) 
eq 6 - i$ for all $i \in \{1, 2, 3, 4, 5\}$. This is the definition of the number of derangements of a specific type,often related to the ménage problem or restricted permutations. For $n=5$,the number of such permutations is given by the formula for the number of permutations with no fixed points on the anti-diagonal,which is $D_n$ for even $n$ and a related value for odd $n$. Specifically,for $n=5$,the number of such permutations is $44$.

Consider a square matrix of order $5$ such that $a_{ij} = 0$ for all $i + j = 6$,where $a_{ij} \in \{0, 1\}$ for all $i, j$. In each row as well as in each column,there is only one non-zero element. Then,the number of such matrices is:

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Notation : $D_n = n! \left( \sum_{i=0}^n \frac{(-1)^i}{i!} \right)$

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