Example 2.3.1 There are three $a$'s, four $b$'s, and two $c$'s. Using these nine letters to form a permutation, if it is required that the same letters cannot all be adjacent in the permutation, how many such permutations are there?
Solution: Let I: the set of all permutations of nine letters,
A: the set of permutations in $I$ where all letters $a$ are adjacent,
B: the set of permutations in $I$ where all letters $b$ are adjacent,
C: the set of permutations in $I$ where all letters $c$ are adjacent.
There are 9! permutations of nine letters, but among them, the 3 $a$s can be interchanged, the 4 $b$s can be interchanged, and the 2 $c$s can be interchanged, all resulting in the same permutation, hence $|I|=\frac{9!}{3!4!2!}=1260$.

Next, calculate $A$. Since the three $A$s are all adjacent, consider the three $A$s as one element, so $|A|=$ $\frac{7!}{4!2!}=105$. Similarly, $|B|=\frac{6!}{3!2!}=60,|C|=\frac{8!}{3!4!}=280$,
$$
\begin{array}{l}
|A B|=\frac{4!}{2!}=12,|A C|=\frac{6!}{4!}=30, \\
|B C|=\frac{5!}{3!}=20,|A B C|=3!=6 .
\end{array}
$$

By the principle of inclusion-exclusion,
$$
\begin{aligned}
|\bar{A} \bar{B} \bar{C}|= & |I|-|A \cup B \cup C|=|I|-|A|-|B|-|C| \\
& +|A B|+|A C|+|B C|-|A B C| \\
= & 1260-105-60-280+12+30+20-6=871 .
\end{aligned}
$$