Example 2 Let $2 \leqslant r<\frac{n}{2} . A=\left\{A_{1}, A_{2}, \cdots, A_{t}\right\}$ be a family of sets, and $\left|A_{i}\right|=r(i=1,2, \cdots, t)$. Find the maximum value of $t$.
Solution: Obviously, when $A_{1}, A_{2}, \cdots, A_{t}$ are all the $r$-element subsets of $X$ containing a fixed element $x$,
$$
t=C_{n-1}^{r-1}.
$$

Erdös-Ko (Ko Zhao) and Rado proved that $C_{n-1}^{r-1}$ is the maximum value of $t$. That is,

Theorem: Let $2 \leqslant r<\frac{n}{2}$. $\mathscr{A}=\left\{A_{1}, A_{2}, \cdots, A_{t}\right\}$ is a family of sets, and $|A|=r$ for $i=1,2, \cdots, t$, then
$$
t \leqslant C_{n-1}^{r-1}.
$$

Equality holds in (2) if and only if $A_{1}, A_{2}, \cdots, A_{t}$ are all the subsets of $X$ containing a fixed element $x$.

This theorem is extremely important in set family theory and is considered a milestone. There are multiple proofs of it, and below we introduce Katona's proof.

We know that $n$ numbers can be arranged on a circle in $(n-1)!$ ways. Similarly, dividing the circle into $n$ equal arcs and labeling each with $n$ numbers $1,2, \cdots, n$ also has $(n-1)!$ methods. Each such arrangement is called a circle of $X=\{1,2, \cdots, n\}$. The key point of Katona's proof is to "embed" $A_{j}$ into the circle.
If the $r$ elements of the $r$-element subset $A_{j}$ are labeled on $r$ consecutive arcs of circle $C$, then these $r$ arcs are called $A_{j}$, and circle $C$ is said to contain subset $A_{j}$. Since the $r$ elements of $A_{j}$ can be arranged in $r!$ ways, and the $n-r$ elements not in $A_{j}$ can be arranged in $(n-r)!$ ways, there are $r!(n-r)!$ circles containing subset $A_{j} (1 \leqslant j \leqslant t)$.
On the other hand, if circle $C$ contains set $A_{1}=\left\{a_{1}, a_{2}, \cdots, a_{r}\right\}$, and $\widehat{P_{1} P_{2}}, \overparen{P_{2} P_{3}}, \cdots, \mathscr{P}_{r} P_{r+1}$ are labeled with numbers $a_{1}, a_{2}, \cdots, a_{r}$ (where $P_{1}, P_{2}, \cdots, P_{n}$ are the equally divided points of the circle), then any other set $A_{j}$ in $C$ shares a common element with $A_{1}$. For each $k (1 \leqslant k \leqslant r+1)$, there are two sets of $r$ consecutive arcs starting from $P_{k}$ (i.e., $\widetilde{P}_{k} P_{k+1}, \widetilde{P_{k+1} P_{k+2}}, \cdots, \widetilde{P}_{k+r-1} P_{k+r}$ and $\left.\widetilde{P}_{k} P_{k-1}, \mathscr{P}_{k-1} P_{k-2}, \cdots, \bar{P}_{k-+1} P_{k-r}\right)$, which share no common points except $P_{k}$, so at most one of them is a set in $\mathscr{A}$ (since any two sets $A_{i}, A_{j} \in \mathscr{A}$ must share a common arc). Since the $r$ consecutive arcs starting from $P_{1}$ or $P_{r+1}$ contain only one set in $\mathscr{A}$, i.e., $A_{1}$. Therefore, each circle $C$ contains at most $r$ sets in $\mathscr{A}$. Combining the above two aspects,
$$
\begin{array}{l}
\sum_{i=1}^{t} \sum_{C \ni A_{i}} 1=t \times r! \times (n-r)! \\
=\sum_{C} \sum_{A_{i} \in C} 1 \leqslant r \times (n-1)!,
\end{array}
$$

i.e.,

Next, we study the case when equality holds.
In the theorem, it is already stated that when $A_{1}, A_{2}, \cdots, A_{t}$ are all the subsets of $X$ containing a fixed element $x$, equality holds in (2).

Conversely, if equality holds in (2), then equality holds in (3), so each circle contains exactly $r$ sets $A_{j}$.
For circle $C$, let the points be $P_{1}, P_{2}, \cdots, P_{n}$, and $\overparen{P_{1} P_{2}}, \overparen{P_{2} P_{3}}, \cdots, \overparen{P_{r} P_{r+1}}$ be labeled with numbers $a_{1}, a_{2}, \cdots, a_{r}, A_{1}=\left\{a_{1}, a_{2}, \cdots, a_{r}\right\}$. According to the above proof, the $r$ consecutive arcs starting from $P_{k} (1 \leqslant k \leqslant r)$ are exactly one set in $\mathscr{A}$, let's say $A_{k}$. There are two cases:
(i) All $A_{k} (1 \leqslant k \leqslant r)$ contain $a_{r}$ (Figure 4.5.2).
(ii) There is an $h (1 \leqslant h < r)$ such that $A_{h}$ contains $a_{r}$ but $A_{h+1}$ does not contain $a_{r}$ (in this case, $A_{h+2}, \cdots, A_{r}$ do not contain $a_{r}$. Figure 4.5.3).

In either case, these $r$ sets (corresponding arcs) cover only $2r-1$ arcs of the circle, not the entire circle (since $n \geqslant 2r$). These $2r-1$ arcs have a starting point (in Figure 4.5.2 it is $P_{1}$, in Figure 4.5.3 it is $P_{n+h-r+1}$) and an endpoint (in Figure 4.5.2 it is $P_{2r}$, in Figure 4.5.3 it is $P_{h+r}$). Without loss of generality, we assume the starting point is $P_{1}$, and the $r$ sets belonging to $\mathscr{A}$ are
$$
\begin{array}{l}
\left\{a_{1}, a_{2}, \cdots, a_{r}\right\},\left\{a_{2}, a_{3}, \cdots, a_{r+1}\right\}, \cdots, \\
\left\{a_{r}, a_{r+1}, \cdots, a_{2r-1}\right\}.
\end{array}
$$

Let the number labeled on $\Gamma_{n} P_{1}$ be $b$, then
$$
B=\left\{b, a_{1}, a_{2}, \cdots, a_{r-1}\right\} \notin \mathscr{A}.
$$

Now we prove that any $r$-element subset $A_{p}$ containing $a_{r}$ and not containing $b$ belongs to $\mathscr{A}$.
Let $A_{0}$ contain $r-s$ numbers in $\left\{a_{1}, a_{2}, \cdots, a_{r}\right\}$. Without loss of generality, assume they are $a_{s+1}, a_{s+2}, \cdots, a_{r}$ (otherwise, renumber $a_{1}, \cdots, a_{r-1}$). Let the other numbers be $c_{r+1}, c_{r+2}, \cdots, c_{r+s}$.
Consider the circle $C^{\prime}$ where the arcs are labeled with $b, a_{1}, a_{2}, \cdots, a_{r}, c_{r+1}, c_{r+2}, \cdots$.
Since $A_{1}=\left\{a_{1}, a_{2}, \cdots, a_{r}\right\} \in \mathscr{A}$ and $B=\left\{b, a_{1}, \cdots, a_{r-1}\right\} \notin \mathscr{A}$, according to the above analysis, case 2 will not appear in circle $C^{\prime}$ (otherwise, it would be equivalent to $A_{n+1}$, and $B \in \mathscr{A}$, meaning case 1 must occur,
$$
\begin{array}{l}
\left\{a_{2}, \cdots, a_{r}, c_{r+1}\right\},\left\{a_{3}, \cdots, a_{r}, c_{r+1}, c_{r+2}\right\}, \cdots, \\
\left\{a_{r}, c_{r+1}, \cdots, c_{2r-1}\right\}
\end{array}
$$

are all subsets in $\mathscr{A}$. In particular, $A_{p} \in \mathscr{A}$.
Furthermore, we prove that any set $A_{q}$ not containing $a_{r}$ must not belong to $\mathscr{A}$.
The complement of $A_{q}$ has $n-r \geqslant r+1$ elements. If these $n-r$ elements include $b$, remove $b$ and then remove some elements to form an $r$-element set containing $a_{r}$. If these $n-r$ elements do not include $b$, remove some elements to form an $r$-element set containing $a_{r}$. According to the above proof, this $r$-element subset containing $a_{r}$ belongs to $\mathscr{A}$. Therefore, $A_{q} \notin \mathscr{A}$.

Finally, by $|\mathscr{A}|=C_{n-1}^{r-1}$ and the fact that sets not containing $a_{r}$ do not belong to $\mathscr{A}$, we conclude that the sets containing $a_{r}$ form
Therefore, there are $n$ ways for $|\mathcal{A}|$ to reach its maximum value.