$36 \cdot 63$ In a tennis tournament, $n$ women and $2 n$ men play, and each player plays against every other player exactly once. If there are no ties, and the ratio of the number of games won by women to the number of games won by men is $7 / 5$, then $n$ is equal to
(A) 2.
(B) 4.
(C) 6.
(D) 7.
(E) None of these.
(29th American High School Mathematics Examination, 1978)
[Solution] Let $k$ denote the number of games won by women against men, and let $W$ and $M$ represent the number of games won by women and men, respectively.

According to the problem, the total number of games, the number of games between one man and one woman, the number of games between two men, and the number of games between two women are
$$
\frac{3 n(3 n-1)}{2}, 2 n^{2}, \quad \frac{2 n(2 n-1)}{2} \text { and } \frac{n(n-1)}{2} \text {, }
$$

Since $k \leqslant 2 n^{2}$, we can gradually deduce that,
$$
\frac{3 n(3 n-1)}{2 k+n(n-1)}=\frac{2(M+W)}{2 W}=\frac{M}{W}+1=\frac{12}{7} \text {, }
$$

which simplifies to
$$
63 n^{2}-21 n=24 k+12 n^{2}-12 n \text {. }
$$

Thus, $\quad \frac{17 n^{2}-3 n}{8}=k \leqslant 2 n^{2}$,
or $17 n^{2}-3 n \leqslant 16 n^{2}$, leading to $n \leqslant 3$.
Substituting $n=1,2,3$ into the equation $\circledast$ yields $k=\frac{7}{4}, \frac{31}{4}, 18$.
Since $k$ must be an integer, then $n=3$.
Therefore, the answer is $(E)$.