7.64 From 1955 given points, what is the maximum number of triplets that can be selected such that any two triplets have 1 point in common?
[Solution] Let point $A$ be a given point and consider the set $M$ of all triples containing point $A$. Since any two triples in $M$ have the common point $A$, the set $M$ satisfies the requirement of the problem and $|M| = C_{1954}^{2}$. Therefore, the maximum value we are looking for is no less than $C_{1954}^{2}$.

On the other hand, suppose there is a set $S$ of triples that satisfies the requirement of the problem and $|S| > C_{1954}^{2}$. Let $\{A, B, C\} \in S$, then any triple in $S$ must contain at least one of the points $A, B, C$. Without loss of generality, assume the triples containing point $A$ are the most. Let $S_A$ be the subset of $S$ consisting of all triples containing point $A$, then
$$
|S_A| \geq \frac{1}{3} C_{1954}^{2} = 977 \times 651 > 325 \times 1953.
$$

In $S_A$, the number of triples containing points $A$ and $B$ is no more than 1953. The number of triples containing points $A$ and $C$ is also no more than 1953. Therefore, there are still triples in $S_A$ that do not contain $B$ and $C$, let $\{A, D, E\}$ be one of them. Using a similar counting method, we can prove that there are still triples in $S_A$ that do not contain $B, C, D, E$, such as $\{A, F, G\}$ and $\{A, H, I\}$, and $\{F, G\} \cap \{H, I\} = \varnothing$. In this way, we have obtained 4 triples:
$$
\{A, B, C\}, \{A, D, E\}, \{A, F, G\}, \{A, H, I\},
$$

and $B, C, D, E, F, G, H, I$ are all distinct. If there is a triple $\{X, Y, Z\} \in S - S_A$, then the triple $\{X, Y, Z\}$ must have a common point with each of the 4 pairs $\{B, C\}, \{D, E\}, \{F, G\}, \{H, I\}$, which is impossible. This means $S = S_A$, i.e., all triples in $S$ contain point $A$. Thus, $S \subset M, |S| < C_{1954}^{2}$; this contradicts the assumption. In conclusion, the maximum number of triples that can be selected to satisfy the requirement of the problem is $C_{1954}^{2}$.