DNF and CNF and Complexity Theory
Counting the number of satisfying assignments to $F$ is at least as hard as determining whether there is a satisfying assignment.  (If the count is 0, there are no satisfying assignments; if the count is $\ge 1$, there is a satisfying assignment.)
For CNF formulas, testing whether there is a satisfying assignment is the SAT problem, which is a classic example of a NP-hard problem.
Counting the number of satisfying assignments of $F$ is at least as hard as determining whether all assignments satisfy $F$.  For DNF formulas, this is NP-hard, too.