The power set of finite set V, has cardinality 2 ^ (cardinality of V), so make sure you heed iMalc's warning.

Bear in mind that this includes the null set and V and itself. My guess is you want the non-trivial, proper subsets of your set?

Code:

suppose A = {2,4,7,13}
Subsets of length n below, using indices 0-3 to describe the set's (array's) original elements:
of length 0:
{} (not too interesting for permuting)
of length 1:
{0} {1} {2} {3} (not to interesting for permuting either)
of length 2:
{0,1} {0,2} {0,3} {1,2} , {1,3} , {2, 3}
... etc
for our particular set A, subsets of length 2 translate to
{2,4} {2, 7} {2, 13} {4, 7} {4, 13}, {7, 13}

I'm not sure if this is the optimal way to approach it, or if the above is easily (non-uglily) implemented. One advantage I can think of with the above is that you won't double-count subsets by counting {2,4} and {4,2}. But I've not tried this problem before. Out of curisoity, what are you working on? Is this just a personal exercise? It's a cool question.