In the book "Combinatorial Foundations" edited by Mei Xiangming ${ }^{[1]}$, there is a combinatorial counting problem: A TV station broadcasts $r$ commercial advertisements over $n$ days. The questions are:
(1) If at least $p$ advertisements are broadcast each day $(n p \leqslant r)$, how many broadcasting methods are there?
(2) If at most $q$ advertisements are broadcast each day $(n q \geqslant r)$, how many broadcasting methods are there?
(3) If the number of advertisements broadcast each day is between $[p, q]$, how many broadcasting methods are there?
First, the key points of the solution in [1] are as follows:
(1) Let the number of advertisements broadcast on the $i$-th day be $x_{i}$, then $(i=1,2, \cdots n)$, we have
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=r, \\
x_{i} \geqslant p(i=1,2, \cdots, n) .
\end{array}
$$

Thus,
$$
\left(x_{1}-p\right)+\left(x_{2}-p\right)+\cdots+\left(x_{n}-p\right)=r-n p \quad x_{i}-p \geqslant 0(i=1,2 \cdots n),
$$

The total number of ways to divide $r-n p$ into $n$ non-negative integers is $\mathrm{C}_{r-n p+n-1}^{n-1}$.
(2) At this point, we have $x_{1}+x_{2}+\cdots+x_{n}=r \quad x_{i} \leqslant q(i=1,2, \cdots, n)$.

Thus,
$$
\begin{array}{c}
\left(q-x_{1}\right)+\left(q-x_{2}\right)+\cdots+\left(q-x_{n}\right)=n q-r, \\
q-x_{i} \geqslant 0(i=1,2, \cdots, n) .
\end{array}
$$

There are $\mathrm{C}_{n q-r+n-1}^{n-1}$ methods.
(3) Let $A=\{n$ days with a total of $r$ advertisements broadcast in various ways $\}$,
$B=\{$ methods of broadcasting at least $p$ times per day $(n p \leqslant r)$ $\}$,
$C=\{$ methods of broadcasting at most $q$ times per day $(n q \geqslant r)$ $\}$,

Then $A=B \cup C, \quad|A|=\mathrm{C}_{r+n-1}^{n-1}, \quad|B|=\mathrm{C}_{r-n p+n-1}^{n-1}, \quad|C|=\mathrm{C}_{m=n}^{n-1}, n+1$. The problem requires
$$
\begin{aligned}
|B \cap C| & =|B \cup C|-|B|-|C| \\
& =|A|-|B|-|C| \\
& =C_{r+n-1}^{n-1}-\mathrm{C}_{r-n p+n-1}^{n-1}-\mathrm{C}_{m-r+n-1}^{n-1} .
\end{aligned}
$$

The underlined parts in the solutions (2) and (3) are incorrect.
(2) $x_{i} \leqslant q$ actually means $0 \leqslant x_{i} \leqslant q$, so $q-x_{i}$ should satisfy $0 \leqslant q-x_{i} \leqslant q$. The discussion is still about the number of non-negative integer solutions with restrictions, while $\mathrm{C}_{n q-r+n-1}^{n-1}$ is the number of ways to express $n q-r$ as $n$ non-negative integers (without restrictions). Therefore, the method in (2) is not $\mathrm{C}_{n q}^{n-1}-r+n-1$.
(3) $A=B \cup C$ is incorrect; only $A \supset B \cup C$ can be obtained.

For example,
$$
x_{1}q, x_{1}+x_{2}+\cdots+x_{n}=r,
$$

then $\left(x_{1}, x_{2}, \cdots, x_{n}\right) \notin B,\left(x_{1}, x_{2}, \cdots, x_{n}\right) \notin C,\left(x_{1}, x_{2}, \cdots, x_{n}\right) \in A$,
and $|B \cap C|=|B|+|C|-|B \cup C|$ is not $|B \cup C|-|B|-|C|$.
Three, the correct solutions for (2) and (3)
(2) Let $B_{i}=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) ; x_{1}+x_{2}+\cdots+x_{n}=r, x_{i}>q, x_{j} \geqslant 0, j \neq i\right\}$,
The number of broadcasting methods sought in (2) is:
$$
\begin{aligned}
\left|\bigcap_{i=1}^{n} \bar{B}_{i}\right|= & |A|-\left|\bigcup_{i=1}^{n} B_{i}\right| \\
= & |A|-\sum_{i=1}^{n}\left|B_{i}\right|+\sum_{1 \leqslant i\frac{r}{q+1}$, $\mathrm{C}_{r-k(q+1)+n-1}^{n-1}=0$,

Therefore,
$$
\left|\bigcap_{i=1}^{n} \bar{B}_{i}\right|=\sum_{k=0}^{\left[\frac{r}{q+1}\right]}(-1)^{k} \mathrm{C}_{n}^{k} \mathrm{C}_{r-k(q+1)+n-1}^{n-1} .
$$
(3) At this point, $x_{1}+x_{2}+\cdots+x_{n}=r, p \leqslant x_{i} \leqslant q \quad(i=1,2, \cdots, n)$, we have $\left(x_{1}\right.$ $-p)+\left(x_{2}-p\right)+\cdots+\left(x_{n}-p\right)=r-n p, 0 \leqslant x_{i}-p \leqslant q-p \quad(i=1,2$, $\cdots, n)$.
From (2), the total number of broadcasting methods for (3) is:
$$
\left[\sum_{k=0}^{\left.\frac{r-n p}{q-p+1}\right]}(-1)^{k} \mathrm{C}_{n}^{k} \mathrm{C}_{(r-n p)-k(q-p+1)+n-1}^{n-1} .\right.
$$