Skip to main content

2008 AMC 10A Problem 23

Problem 23 of 25HarderCombinatorics

Two subsets of the set S={a,b,c,d,e}S = \{a, b, c, d, e\} are to be chosen so that their union is SS and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?

Answer choices

Show solution

Solution

Choose the two common elements in (52)=10\binom{5}{2} = 10 ways. Each of the remaining 33 elements must lie in exactly one subset, giving 23=82^3 = 8 assignments, for 8080 ordered pairs. Since the order of the two subsets does not matter, divide by 22 to get 802=40.\dfrac{80}{2} = 40. Thus, the correct answer is B.
AoPS wiki

Tagged: subsets · combinations · multiplication principle

More practice