2008 AMC 10A Problem 23

Below is the professionally curated solution for Problem 23 of the 2008 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10A solutions, or check the answer key.

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Concepts:subsets combinations multiplication principle

Difficulty rating: 1770

23.

Two subsets of the set $S = \{a, b, c, d, e\}$ are to be chosen so that their union is $S$ 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?

$20$

$40$

$60$

$160$

$320$

Solution:

Choose the two common elements in $\binom{5}{2} = 10$ ways.

Each of the remaining $3$ elements must lie in exactly one subset, giving $2^3 = 8$ assignments, for $80$ ordered pairs.

Since the order of the two subsets does not matter, divide by $2$ to get $\dfrac{80}{2} = 40.$

Thus, the correct answer is B.

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