This seems as if it should be an easy question, but I am stuck. An equivalent way of thinking about it is to view c as a vector of the exponents of its prime factors and to determine how many ways there are of creating vectors a and b such that the maximum of the ith components of a and b is equal to the ith component of c.
I think I can get a lower bound by considering the cases where, for each exponent value in c, one of the corresponding vector components of a and b equals the vector component exp of c and the other vector has a component no greater than exp - 1. The number of such cases is $2^{(k-1)}$ times the product of the exp where k is the size of the vectors and exp is the value in c, keeping in mind that the total number of possibilities from 0 to exp - 1 is exp. The $2^{(k-1)}$ comes from the number of ways of choosing either a or b to contain exp and dividing by 2 to remove the redundancy between a and b. Allowing both a and b to have a value of exp creates redundancy issues that I don't know how to handle.
After looking at mathlove's references and seeing how simple the fleablood's equation is, I was encouraged to generalize the question. How many ordered sets of numbers of size k can you find such that the LCM of each equals c? It seems that a fairly straightforward use of the inclusion/exclusion principle can be used for each different prime in c, and then multiply those numbers together. I have no idea how to then find the number of unordered sets of k numbers.