Number of $n\times n$ non-singular $0-1$ matrices Problem: Find the number of $n\times n$ non-singular matrices where an entry is either $0$ or $1$.
PS: This was motivated by 779412
 A: The answer corresponds to OEIS sequence A055165. Little seems to be known apart from the forst eight values:
The proportion of singular matrices (among the $2^{n^2}$ candidates) is conjectured to be ${n+1\choose 2}\frac 1{2^{n-1}}$ 
A: Let $\mathbb{F}_q$ a finite field with $q$ elements. Let $G=\{g=(a_{ij})_{i,j \le n}  \ ; a_{ij} \in \mathbb{F}_q, \det g \neq 0 \} $. So, we have 
$$|G|=q^{n(n-1)/2} \cdot (q^n-1)(q^{n-1}-1)\cdots(q-1)$$
We proceed by induction: if n=1, is trivial.
Let assume $|GL(n-1,q)|=q^{(n-1)(n-2)/2} \cdot (q^{n-1}-1)(q^{n-1}-1)\cdots(q-1)$
Let $V=\mathbb{F}_q^n$. We know that $G$ is transitive about $V- \{0\}$. By Orbit-stabilizer Theorem, $|G|=|V- \{0\}| \cdot G_{b_1}$, where $b_1=(1,0,...,0)$ and $G_{b_1}$ is the stabilizer of $b_1$ . Notice that $|G_{b_1}|=|GL(n-1,q)| \cdot q^{n-1}$. Then
$$|G|=(q^n-1) \cdot q^{(n-1)(n-2)/2} \cdot (q^{n-1}-1)(q^{n-1}-1)\cdots(q-1) \cdot q^{n-1}$$
$$|G|=q^{n(n-1)/2} \cdot (q^n-1)(q^{n-1}-1)\cdots(q-1)$$
I do not know if this helps. How is related, maybe it helps you.
