How many ways are there to arrange $n$ points in the finite field plane $(\Bbb F_q)^2$ with no three of the points collinear?
An easy upper bound is $(q^2)^n=q^{2n}$, but of course it's less than that. (Of course, if I asked the same question over $\Bbb R^2$, it would be infinite.)
The collinearity condition can be expressed as follows: $(a_0,b_0)$, $(a_1,b_1)$, and $(a_2,b_2)$ are collinear iff: $$\begin{vmatrix}1&1&1\\a_0&a_1&a_2\\b_0&b_1&b_2\end{vmatrix}=0$$ Thus the question makes sense over a finite field.
The way to do this is probably some inclusion-exclusion argument (or perhaps some fancier poset Möbius argument) but I'm unsure on the details.
I hope, perhaps unreasonably, that for fixed $n$, the answer is a polynomial in $q$.