6. In the third year of a certain school, there are four classes, each class selects one male and one female table tennis player to form a mixed doubles team for the grade. Then, among the four mixed doubles pairs, the probability that no pair of players are from the same class is ( ).
(A) $\frac{5}{12}$
(B) $\frac{4}{9}$
(C) $\frac{3}{8}$
(D) $\frac{2}{5}$
6. C.
4 men and 4 women forming four mixed doubles pairs have a total of 24 combinations.

Let the male and female players from classes one, two, three, and four be $A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3}$, and $A_{4}, B_{4}$. Then, the scenarios where no pair of players are from the same class are as follows:
$$
\begin{array}{l}
A_{1} A_{2} A_{3} A_{4} A_{1} A_{2} A_{3} A_{4} \quad A_{1} A_{2} A_{3} A_{4} \\
\text { | | | |, | | | |, | | | |, } \\
B_{2} B_{1} B_{4} B_{3} \quad B_{2} B_{3} B_{4} B_{1} \quad B_{2} B_{4} B_{1} B_{3} \\
A_{1} A_{2} A_{3} A_{4} \quad A_{1} A_{2} A_{3} A_{4} \quad A_{1} A_{2} A_{3} A_{4} \\
\text { | | | |, | | | |, | | | |, } \\
B_{3} B_{1} B_{4} B_{2} \quad B_{3} B_{4} B_{1} B_{2} B_{3} B_{4} B_{2} B_{1} \\
A_{1} A_{2} A_{3} A_{4} \quad A_{1} A_{2} A_{3} A_{4} \quad A_{1} A_{2} A_{3} A_{4} \\
\text { | | | |, | | | |, | | | |. } \\
B_{4} B_{3} B_{2} B_{1} \quad B_{4} B_{3} B_{1} B_{2} B_{4} B_{1} B_{2} B_{3} \\
\end{array}
$$

From the above, there are a total of 9 scenarios. Therefore, the probability that in four mixed doubles pairs, no pair of players are from the same class is $\frac{3}{8}$.