$4 \cdot 7$ Let $P_{1}, P_{2}, \cdots, P_{n}$ be $n$ points in the plane, where no three points are collinear. Connect $m$ line segments between them such that among any 4 points, there are 3 points that form a triangle, i.e., every two points among them are connected by a line segment. Find the minimum value of $m$.


[Solution] When any two points among $P_{1}, P_{2}, \cdots, P_{n-1}$ are connected by 1 line, a total of $C_{n-1}^{2}$ lines are connected, and this method of connection clearly meets the requirements of the problem, so the minimum value sought is no greater than $C_{n-1}^{2}$.

On the other hand, suppose there is no line segment between points $P_{1}, P_{2}$. Then any two points among the remaining $n-2$ points must be connected by a line. Otherwise, there must be $3 \leqslant i<j \leqslant n$ such that points $P_{i}$ and $P_{j}$ are not connected by a line, and thus the 4 points $\left\{P_{1}, P_{2}, P_{i}, P_{j}\right\}$ do not meet the requirements of the problem, leading to a contradiction. In addition, for any $3 \leqslant i \leqslant n$, point $P_{i}$ must be connected by a line to at least one of $P_{1}, P_{2}$. Otherwise, it will again lead to 4 points not meeting the requirements of the problem. Thus, the number of lines in a set of connections that meets the requirements of the problem is no less than
$$
C_{n-2}^{2}+n-2=C_{n-1}^{2} \text {. }
$$

In summary, the minimum value of $m$ is $C_{n-1}^{2}$.