Example 6. A person wrote 6 letters to 6 different people and prepared 6 envelopes with the recipients' addresses written on them. How many ways are there to place the letters into the envelopes so that no letter matches the recipient on the envelope? (Polish Competition Question)
Let's denote the set of all ways to place 6 letters into 5 different envelopes (i, $2, \cdots, 6$) such that one letter is left out, and the remaining letters are placed into the 5 envelopes in any order, as $\Lambda_{i}$. Then, $\left|A_{i}\right|=P_{5}^{5}=5!$.
Similarly,
$$
\begin{array}{l}
\left|A_{i} \cap A_{j}\right|=P_{4}^{4}=4!, \\
\cdots \cdots \cdots \\
\left|A_{1} \cap A_{2} \cap A_{3} \cap A_{4} \cap A_{5} \cap A_{6}\right|=P_{1}^{1}=1!.
\end{array}
$$

The scenario where none of the 6 letters match the recipient on the envelopes is $\overline{A_{1}} \cap \overline{A_{2}} \cap \cdots \cap \overline{A_{6}}$.
By the principle of inclusion-exclusion,
$$
\begin{array}{l}
\left|\overline{A_{1}} \cap \overline{\Lambda_{2}} \cap \cdots \cap \overline{A_{5}}\right| \\
=|S|-\sum_{1 \leq i \leq 6}\left|A_{i}\right|+\sum_{1 \leq i<j \leq 6}\left|A_{1} \cap A_{1}\right|-\cdots+ \\
(-1)^{6}\left|A_{1} \cap A_{2} \cap A_{3} \cap A_{4} \cap A_{5} \cap A_{5}\right| \\
=6!-C_{6}^{1} \cdot 5!+C_{6}^{2} \cdot 4!-C_{6}^{3} \cdot 31 \\
+C_{6}^{4} \cdot 2!-C_{6}^{5} \cdot 1!+C_{6}^{5} \\
=265 .
\end{array}
$$

Thus, the number of ways to place the letters such that none of the letters match the recipient on the envelopes is 265.