permutations and combinations

Question

A student is given a reading list of ten book from which he must select two for an outside reading requirement. In how many ways can he make his selection?

Answer 1

Given a list of ten books, from which he must choose two. This is the start of a combination / permutation problem. We must now determine whether this is a combination (order doesn’t matter (AB is the same as BA) or a permutation (order matters (AB is different than BA)). In this case order doesn’t matter because reading book A than book B is the same as reading book B than book A. We now use the formula for combinations: ₁₀C₂ = 10! / ((10-2)! * 2!) = 10! / (8! * 2!) = (10*9) / 2 = 45 ways.