Question

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?

$56$

$58$

$60$

$62$

$64$

Answer: D

Concepts:multiplication principle complementary counting

Difficulty rating: 1330

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.

2007 AMC 10A Problem 12

Answer: D

Difficulty rating: 1330

Solution:

Problem 12 in Other Years