$5 . n$ tennis players $(n \geqslant 4)$, any two of them can form a doubles team. The number of ways to select two teams from them is ( ).
(A) $\mathrm{C}_{n}^{4}$ ways
(B) $6 \mathrm{C}_{n}^{4}$ ways
(C) $3 C_{n+1}^{4}$ ways
(D) $\left(6 C_{n}^{4}+3 C_{n}^{3}\right)$ ways
5. (C).

Solution 1: For $n$ athletes, the number of pairs is $\frac{n(n-1)}{2}$. Selecting 2 pairs from these, we have
$$
\begin{aligned}
\frac{C_{n(n-1)}^{2}}{2} & =\frac{1}{2} \cdot \frac{n(n-1)}{2} \cdot\left(\frac{n(n-1)}{2}-1\right) \\
& =\frac{1}{8}(n+1) n(n-1)(n-2) \\
& =3 C_{n+1}^{+} .
\end{aligned}
$$

Solution 2: The composition of the two pairs of players taken out could be 4 people $(a, b, c, d)$, or 3 people $(a, b, c)$. Selecting 4 people and then dividing them into two groups has $3 \mathrm{C}_{n}^{4}$ ways, and selecting 3 people and then dividing them into two groups has $3 \mathrm{C}_{n}^{3}$ ways. In total, there are $3 \mathrm{C}_{n}^{4}+3 \mathrm{C}_{n}^{3}=3 \mathrm{C}_{n+1}^{4}$.