2007 AMC 10A Problem 12

Below is the professionally curated solution for Problem 12 of the 2007 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10A solutions, or check the answer key.

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Concepts:multiplication principle complementary counting

Difficulty rating: 1330

12.

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?

$56$

$58$

$60$

$62$

$64$

Solution:

Each tourist independently chooses one of the two guides, giving $2^6 = 64$ arrangements.

Exactly two of these leave a guide with no tourists, so the answer is $64 - 2 = 62.$

Thus, the correct answer is D.

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