Given a subset of size 3, how many cycles can we form? There are 3! total orderings of the elements, but each cycle is counted 3 times because there are 3 different places to start listing the cycle. So the total number of cycles is 3! / 3.
Similarly, we can form 5! / 5 different cycles from a subset of size 5.
So all together, there are
permutations of the specified type.