How many ways can you choose 2 letters from {A, B, C, D}, where order doesn't matter? Using the formula: 4C2 = 4! / (2! × 2!) = 24 / 4 = 6. Listing them confirms it: {A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C,D} — six unordered pairs, half as many as the twelve permutations of the same set.