2004 AMC 10A Problem 13

Think about the list of every (man, woman) pair who danced together. Count that list two different ways.

1. Counting by the men. There are $12$ men, and each is in $3$ pairs:
$$12 \times 3 = 36 \text{ pairs}.$$
2. Counting by the women. Each woman is in $2$ pairs. If there are $w$ women, the list has $2w$ pairs.
3. It is the same list, so the two counts agree:
$$2w = 36, \qquad w = 18.$$

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

This is called double counting: one collection (the dance pairs) is counted once from each side, and both counts must give the same number. The same trick works for handshakes, games in a tournament, and people on committees.

Use a small group. $2$ men make $2 \times 3 = 6$ pairs, and $3$ women also make $3 \times 2 = 6$ pairs. So men and women come in the ratio $2 : 3$. With $12$ men, that is $18$ women.

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 $\boxed{\textbf{(D)}\ 18}$.