This question is inspired from [here](https://mathoverflow.net/questions/18547/number-of-unique-determinants-for-an-nxn-0-1-matrix), where it was asked what possible determinants an $n \times n$ matrix with entries in {0,1} can have over $\mathbb{R}$. 


My question is: how many such matrices have non-zero determinant? 


If we instead view the matrix as over $\mathbb{F}\_2$ instead of $\mathbb{R}$, then the answer is 


$(2^n-1)(2^n-2)(2^n-2^2) \dots (2^n-2^{n-1}).$


This formula generalizes to all finite fields $\mathbb{F}\_q$, which leads us to the more general question of how many $n \times n$ matrices with entries in { $0, \dots, q-1$ } have non-zero determinant over $\mathbb{R}$? 



Number of invertible {0,1} real matrices?
See [Sloane, A046747](http://oeis.org/A046747) for the number of singular (0,1)-matrices. It doesn't seem like there's an exact formula, but it's conjectured that the probability that a random (0,1)-matrix is singular is asymptotic to $n^2/2^n$.


Over $F\_2$ the probability that a random matrix is nonsingular, as $n \to \infty$, approaches the product $(1/2)(3/4)(7/8)\cdots = 0.2887880951$, and so the probability that a random large matrix is singular is only around 71 percent. I should note that a matrix is singular over $F\_2$ if its real determinant is even, so this tells us that determinants of 0-1 matrices are more likely to be even than odd. 


