If $n$ is odd, then is the number of $0-1$ matrices whose determinant is odd, almost always an odd value?
Above seems to be wrong:
The number of non-singular determinants over $\Bbb F_2$ is even. Does this indicate parity of number of odd determinants or permanents over $\Bbb R$ is even?
$$\#\operatorname{GL}(n,\mathbb{F}_2)=(2^n-1)(2^n-2)(2^n-4)\cdot\ldots\cdot(2^n-2^{n-1})$$ hence: $$\nu_2\left(\#\operatorname{GL}(n,\mathbb{F}_2)\right) = 1+2+\ldots+(n-1)=\frac{n(n-1)}{2}$$ and the number of $0-1$ matrices with odd determinant (i.e. invertible) is a multiple of $2^{\binom{n}{2}}$.