The Number of different $n \times n$ skew-symmetric matrices with each element being equal to either $0$ or $1$ or $-1$ where $n=5$ is ?
The Number of different $n \times n$ Skew-symmetric matrices with each element being equal to either $0$ or $1$ or $-1$
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matrices
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0You just changed the question altogether and I had posted an answer addressing the previous case. Also your edited question is a duplicate. What is going on??? – 2017-01-05
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0@WiCK3DPOiSON: Not really a duplicate: this question asks about *skew-symmetric* matrices (the original version asked about *symmetric* ones), while the one that you claim this one to be a duplicate of asks only about *arbitrary* matrices. – 2017-01-05
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Given $n\times n$ matrix with restricted $m$ elements and matrix is skew-symmetric, then number of possible matrices is $$m^{(\frac{(n)(n-1)}{2})}$$ since diagonal elements are 0 and we have to fill upper or lower triangular part of the matrix. The rest elements will be automatically be taken care due to (skew-)symmetry.
In this case both $n=5$ and $m=3$, therefore $3^{10}$ matrices.