Example 5 In a local area network consisting of several computers, for any $m(m \geqslant 3)$ computers there is a unique one that is connected to all of them. How many computers at most can be connected to the same computer in this local area network?
Analysis: We should first intuitively guess that the answer to this problem is $k = m$. To clarify the problem, it is best to use proof by contradiction.

Solution: Let the maximum number of computers that can be connected to the same computer be $k$. Without loss of generality, assume that computers $B_{1}, B_{2}, \cdots, B_{k}$ are all connected to computer $A$, and let $S=\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$.

Clearly, $k \geqslant m$. If $k > m$, let $\left\{B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}\right\}$ be any $(m-1)$-element subset of $S$, then $A, B_{i_{1}}, \cdots, B_{i_{m-1}}$ these $m$ computers are all connected to some computer, denoted as $C_{i}$.
Since $C_{i}$ is connected to $A$, it follows that $C_{i} \in S$. Define the mapping:
$$
f:\left\{B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}\right\} \rightarrow C_{i} \in S .
$$

Then $f$ is an injection from the set of all $(m-1)$-element subsets of $S$ to $S$. In fact, if there are two different $(m-1)$-element subsets $\left\{B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}\right\}$ and $\left\{B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{m-1}}\right\}$ of $S$ that both correspond to $C_{i}$, considering that $\left\{B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}\right\} \cup \left\{B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{m-1}}\right\}$ contains at least $m$ elements, these $m$ computers are all connected to both $A$ and $C_{i}$, which contradicts the given conditions.

Since $f$ is an injection, we have $\mathrm{C}_{m}^{k-1} \leqslant k$. On the other hand, since $m \geqslant 3, m-1 \geqslant 2$, and $k > m$, it follows that $\mathrm{C}_{k}^{m-1} > \mathrm{C}_{k}^{1} = k$, which is a contradiction. Therefore, the maximum value sought is $m$.