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2007 AMC 10A Problem 12

Problem 12 of 25IntermediateCounting & Probability

Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?

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Solution

Each tourist independently chooses one of the two guides, giving 26=642^6 = 64 arrangements. Exactly two of these leave a guide with no tourists, so the answer is 64−2=62.64 - 2 = 62. Thus, the correct answer is D.

More practice

Concepts: multiplication principle · complementary counting

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.