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2004 AMC 10A Problem 13

2004 AMC 10A Problem 13AMC 10 Step by Step

Name: Date: September 13, 2026

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

At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

A)
$8$
B)
$12$
C)
$16$
D)
$18$
E)
$24$

Answer

Key insight

Count man-woman dance pairs two ways: 12 men times 3 equals (number of women) times 2, so there are 18 women.

Solution

Consider the set of all (man, woman) pairs who danced together, and count it from each side.

From the men's side: $12$ men, each in $3$ pairs, so there are $12 \cdot 3 = 36$ pairs.

From the women's side: if there are $w$ women, each in $2$ pairs, the same set has $2w$ pairs.

Equating, $2w = 36$ , so $w = 18$ .

The answer is $\boxed{\textbf{(D)}\ 18}$ .

Why this works

This is double counting: a single collection of "incidences" (here, dance partnerships) is tallied once by men and once by women, and the two tallies must agree. The same idea handles handshakes, edges of a graph, and seats in committees. The number of pairs is the bridge between the two unknown-related quantities.

Alternative approach

Ratio view: every $2$ men account for $6$ partnerships, which is what $3$ women account for, so men to women is $2 : 3$ . Scaling $12$ men gives $18$ women.

The trap

Inverting the ratio and computing 12 times 2 divided by 3 = 8, choice (A).

Common mistakes

Inverting the ratio and computing 12 times 2 divided by 3 = 8, choice (A).
Assuming the men and women must be equal in number ( $12$ ) because "everyone danced."

Techniques

Map the objects to something easier to count