For example, $12 n$ students go out for a walk every day, in pairs. If every pair of students walks together at most once, how many days can such walks last?
Solution: Since each person has $2n-1$ classmates, the walk can last at most $2n-1$ days. We will prove that with proper arrangement, it can indeed last for $2n-1$ days.

To do this, we construct a graph, using $0, 1, \cdots, 2n-1$ to represent the $2n$ students. The first walk is represented by lines, i.e., the edges in the graph are
$$
\{0,1\},\{2,2n-1\},\{3,2n-2\}, \cdots,\{n, n+1\}.
$$

Then we rotate around $O$, each time by an angle of $\frac{2\pi}{2n-1}$, thus obtaining the arrangements for $2n-1$ walks (for example, the second walk is $\{0,2\},\{1,3\}$, $\{2n-1,4\}, \cdots,\{n+2, n+1\}$).
For $n$ points, connecting each pair of points with an edge results in a complete graph $K_{n}$. Example 1 shows that the $C_{2n}^{2}$ edges of the complete graph $K_{2n}$ can be divided into $2n-1$ groups, each containing $n$ edges, and these $n$ edges have no common (end) points. Such a set of edges is called a 1-factor (1 indicates that each point is connected to only one edge, i.e., each element belongs to only one binary subset) or a perfect matching.
Now we will complete the remaining part of Example 3 in Section 3.7.