$36 \cdot 72$ In a singles table tennis tournament, it was originally planned that each pair of players would play exactly one match. However, after three players each played 2 matches, they withdrew from the tournament. As a result, only 50 matches were played in total. The number of matches played among these three players is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(China High School Mathematics League, 1999)
［Solution］Let the number of matches among the three players be $r$, with a total of $n$ players participating.
According to the problem, we have $50=C_{n-3}^{2}+r+(6-2r)$,
which simplifies to $\frac{(n-3)(n-4)}{2}=44+r$.
Given that $0 \leqslant r \leqslant 3$, upon inspection, it is found that only when $r=1$, $n=13$ is a positive integer. Therefore, the answer is $(B)$.