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?
$\mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24$
If each man danced with $3$ women, then there will be a total of $3\times12=36$ pairs of men and women.  However, each woman only danced with $2$ men, so there must have been $\frac{36}2 \Longrightarrow \boxed{\mathrm{(D)}\ 18}$ women.