I just heard the term skew symmetric matrix and upon discovering what it was, I thought to myself, "Jeez, there could only be so many of those."
I'm not good with the whole permutation thing and this is a new concept to me, I don't know the rules. So, I convey this question to you.
How many skew-symmetric matrices of order $ m \times n$ are possible?
On
If you're looking at $n\times n$ matrices over a finite ring of order $q$, then you are free to pick the entries above the diagonal to be whatever you want. The rest are then determined, and you get a skew-symmetric matrix.
There are $n^2$ total entries, $n^2-n$ nondiagonal entries, and so $(n^2-n)/2$ entries above the diagonal. There are $q$ choices to be made in each spot, so that gives you a total of $q^{\frac{n^2-n}{2}}$ skew symmetric matrices over the finite ring.
If the ring you have is infinite, then you can still produce skew-symmetric matrices the same way, and you'll have infinitely many.
Well, knowing the upper half fixes the lower half so you can pick values for the upper half and the main diagonal. There is an uncountable infinity of those, even for $2 \times 2$ matrices, because you can pick any of the values in an uncountable infinity of choices, and you got more than 1 of those values to pick :-).