2008 AMC 10A Problem 23

Call the subsets $X$ and $Y$. Because $X \cup Y = S$, every element of $S$ lies in at least one of them, so each element is in exactly one of three regions: $X$ only, $Y$ only, or both.

First count ordered pairs $(X, Y)$. Exactly two elements are in both: $\binom{5}{2} = 10$ ways to pick them. Each of the remaining three elements is in $X$ only or in $Y$ only: $2^3 = 8$ ways. That gives $10 \cdot 8 = 80$ ordered pairs.

Since order does not matter, each unordered pair $\{X, Y\}$ has been counted twice, once as $(X, Y)$ and once as $(Y, X)$, provided $X \ne Y$. And $X \ne Y$ always holds here: the three non-shared elements belong to exactly one of the two subsets, so the subsets differ. Therefore the number of unordered pairs is
$$\frac{80}{2} = 40.$$

The answer is $\boxed{\textbf{(B)}\ 40}$.

A pair of subsets covering $S$ is the same thing as a labeling of each element by its Venn region, so "count pairs of sets" becomes "count labelings," a multiplication-principle problem. The union condition removes the "neither" region; the intersection condition fixes how many get the "both" label. The final step, converting ordered to unordered, is safe only after checking that no pair is its own swap.

Fix the two shared elements ($10$ ways). Then $X$ is determined by which of the other three elements it takes, and $Y$ takes the rest; choosing $X$'s share as a subset of $3$ elements gives $8$ ordered choices, or $4$ unordered splits (a split and its complement are the same pair). Total $10 \cdot 4 = 40$.

• Forgetting to divide by 2 for unordered pairs (answering 80, not listed, and then guessing) or halving when the two subsets could be equal; here they never are.
• Allowing an element to be in neither subset (three choices plus "neither" per element), which violates the union condition.