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2007 AMC 10A Problem 12

2007 AMC 10A Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 2 of 5 · about 2 min · position P11-15

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?

A)
$56$
B)
$58$
C)
$60$
D)
$62$
E)
$64$

Answer

Key insight

Each tourist independently picks one of two guides, 2^6 ways; subtract the 2 assignments in which one guide gets nobody.

Solution

Ignore the restriction first. Each of the six tourists has two choices of guide, and the choices are independent, so there are $2^6 = 64$ ways to assign tourists to guides.

Exactly two of those assignments violate the rule: all six choose the first guide, or all six choose the second. Remove them:
$64 - 2 = 62.$

The answer is $\boxed{\textbf{(D)}\ 62}$ .

Why this works

"At least one in each group" is a classic signal for complementary counting: the unrestricted count is a clean power of $2$ , and the forbidden cases are few and easy to list. Note the guides are distinct, so sending everyone to guide A is different from sending everyone to guide B; that is why $2$ , not $1$ , is subtracted.

Alternative approach

Sum over how many tourists go with the first guide: $\binom61 + \binom62 + \binom63 + \binom64 + \binom65 = 6 + 15 + 20 + 15 + 6 = 62$ .

The trap

Dividing by 2 because the two groups look interchangeable, but the guides are distinct people so the groupings are ordered.

Common mistakes

Dividing by $2$ because the two groups look interchangeable, but the guides are distinct people so the groupings are ordered.
Forgetting to subtract the empty-group cases and answering $64$ , or subtracting only one of them and getting $63$ .

Techniques

Count the complement and subtract from the total