【Example 2】Using the digits $1,2,3,4,5$, how many five-digit numbers greater than 20000 can be formed, with no repeated digits and the hundreds place not being the digit 3?
Solve: Since $1,2,3,4,5$ can form $A_{5}^{5}$ five-digit numbers without repeated digits. But because $\left|B_{1}\right|=\left|B_{2}\right|=A_{4}^{4},\left|B_{1} \cap B_{2}\right|=A_{3}^{3}$, the number of five-digit numbers that meet the conditions is
$$
A_{5}^{5}-2 A_{4}^{4}+A_{3}^{3}=65 \text{ (numbers) }
$$

From the above discussion, we understand that for $k=2$, students are not unfamiliar with the inclusion-exclusion principle. Then, what form will the inclusion-exclusion principle take when $k \geqslant 3$?
By observing the Venn diagram in Figure 11, it is not difficult to see that when $k=3$, we have
$$
\begin{array}{l}
\left|B_{1} \cup B_{2} \cup B_{3}\right| \\
=\quad\left|B_{1}\right|+\left|B_{2}\right|+\left|B_{3}\right|-\left|B_{1} \cap B_{2}\right| \\
\quad-\left|B_{2} \cap B_{3}\right|-\left|B_{3} \cap B_{1}\right|+\left|B_{1} \cap B_{2} \cap B_{3}\right|
\end{array}
$$

Formula (2) is the inclusion-exclusion principle for the union of three sets. But what form will it take when $k=4,5,6, \cdots, n$?
To facilitate observation and writing, let's introduce some notation. Let
$$
\begin{array}{l}
S_{1}=\left|B_{1}\right|+\left|B_{2}\right|+\cdots+\left|B_{k}\right| \\
S_{2}=\left|B_{1} \cap B_{2}\right|+\left|B_{1} \cap B_{3}\right|+\cdots+\left|B_{k-1} \cap B_{k}\right| \\
\cdots \\
S_{k}=\left|B_{1} \cap B_{2} \cap \cdots \cap B_{k}\right|
\end{array}
$$

This means that $S_{1}$ represents the sum of the number of elements in each of the sets $B_{1}, B_{2}, \cdots, B_{k}$; $S_{2}$ represents the sum of the number of elements in the intersections of every two sets among $B_{1}, B_{2}, \cdots, B_{k}$; $S_{3}$ represents the sum of the number of elements in the intersections of every three sets; $\cdots \cdots$; and $S_{k}$ represents the number of elements in the intersection of all these sets. From formulas (1) and (2), we know that for $k=2$ and 3, we have
$$
\begin{array}{l}
\left|B_{1} \cup B_{2}\right|=S_{1}-S_{2} \\
\left|B_{1} \cup B_{2} \cup B_{3}\right|=S_{1}-S_{2}+S_{3}
\end{array}
$$

Therefore, we can infer that for $k=n$, we should have
$$
\left|B_{1} \cup B_{2} \cup \cdots \cup B_{3}\right|=S_{1}-S_{2}+\cdots+(-1)^{n-1} S_{n}
$$

This inference is entirely correct, and the inferred result can be rigorously proven (for example, by mathematical induction). Formula (3) is called the inclusion-exclusion principle, and the counting principle using the inclusion-exclusion principle is called the inclusion-exclusion principle. Let's look at a few examples.