6. In a certain singles table tennis tournament, it was originally planned that each pair of players would play one match, but after three players each played two matches, they withdrew from the tournament, resulting in a total of 50 matches being played. The number of matches played among these three players is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
6. B.

Let the number of matches played among these three players be $r$, with a total of $n$ players participating.
According to the problem, we have
$$
\begin{array}{l}
\mathrm{C}_{n-3}^{2}+(6-r)=50 \\
\Rightarrow \frac{(n-3)(n-4)}{2}=44+r .
\end{array}
$$

Since $0 \leqslant r \leqslant 3$, we will discuss four cases.
(1) When $r=0$,
$$
(n-3)(n-4)=88 \Rightarrow n^{2}-7 n-76=0 \text {, }
$$

there is no positive integer solution;
(2) When $r=1$,
$$
(n-3)(n-4)=90 \Rightarrow n=13 \Rightarrow r=1 \text {, }
$$

which meets the problem's requirements;
(3) When $r=2$,
$$
(n-3)(n-4)=92 \Rightarrow n^{2}-7 n-80=0 \text {, }
$$

there is no positive integer solution;
(4) When $r=3$,
$$
(n-3)(n-4)=94 \Rightarrow n^{2}-7 n-82=0 \text {, }
$$

there is no positive integer solution.