Choosing 2 items from 5 gives 10 combinations and 20 permutations. Why is the permutation count exactly twice the combination count here, and what would the factor be for choosing 3?

C. Because each combination can be ordered in 2! ways, counted separately.

The pair {A, B} corresponds to AB and BA, so permutations count it twice; for a set of three there are 3! = 6 arrangements. Read P(n,k) = C(n,k)·k! as a combination being a permutation with the ordering forgotten. Both are defined without repetition, deciding whether order matters and whether repetition is allowed is most of the difficulty in elementary counting.