How many 3-SAT expressions with up to N variables are satisfiable?
There is almost certainly no non-trivial formula that will tell you the number of solutions. By non-trivial, I mean something not of the form
$$
\sum_{x_1, \dots, x_n\in\{0,1\}} f(x_1, \dots, x_n)\,,
$$
where
$$f(x_1, \dots, x_n)=\begin{cases}1 &amp;\text{if $x_1, \dots, x_n$ satisfies the formula}\\
0&amp;\text{otherwise.}\end{cases}
$$

Note that if we have a formula&nbsp;$F$ (trivial or otherwise) that tells us how many satisfying assignments there are, we can decide 3-SAT just by computing&nbsp;$F$ and checking whether $F=0$ (unsatisfiable) or $F&gt;0$ (satisfiable). This means that computing&nbsp;$F$ is NP-hard so, assuming that P$\,\neq\,$NP, $F$&nbsp;is difficult to compute, which suggests that $F$&nbsp;can't have any nice form.

In fact, $F$&nbsp;is complete for the complexity class #P, which is essentially at least as hard as all of the polynomial hierarchy. This means that it seems to be quite a lot harder than "just NP-hard". Furthermore, we can't even approximately compute&nbsp;$F$ in any efficient way unless P=NP. Even being able to approximate&nbsp;$F$ with exponential additive error would allow us to distinguish between $F=0$ and $F&gt;0$ and thus allow us to solve 3-SAT. For example, given a formula with $N$&nbsp;variables and $F$&nbsp;solutions, we can create a formula with $2N$&nbsp;variables and exactly the same clauses (so variables $N+1, \dots, 2N$ aren't mentioned in the formula and can take any value). This formula has $2^NF$ satisfying assignments and being able to approximate that number strictly better than $\pm 2^{N-1}$ would allow us to tell the difference between $F=0$ and $F&gt;0$.