Example 1 A class has 6 lessons scheduled on Monday: Chinese, Math, History, PE, Music, and Geography.
(1) If PE cannot be scheduled as the first lesson in the morning, History cannot be the second lesson in the morning, Music cannot be the third lesson in the morning, Chinese cannot be the fourth lesson in the morning, Math cannot be the first lesson in the afternoon, and Geography cannot be the second lesson in the afternoon, how many different schedules are there?
(2) In question (1), if the condition “Music cannot be the third lesson in the morning” is changed to “Music must be the third lesson in the morning,” how many different schedules are there?
(3) In question (1), if the condition “Geography cannot be the second lesson in the afternoon” is removed, how many different schedules are there?
(4) In question (1), if the restrictions “Math cannot be the first lesson in the afternoon, Geography cannot be the second lesson in the afternoon” are removed, how many different schedules are there?
(5) In question (1), if the condition “Music cannot be the third lesson in the morning” is changed to “Music must be the third or fourth lesson in the morning,” how many different schedules are there?
(6) Chinese, Math, History, and Geography must be scheduled in the morning, the first lesson cannot be History, the second lesson cannot be Geography, the third lesson cannot be Chinese, and the fourth lesson cannot be Math. How many different schedules are there?
(7) Chinese, Math, and History must be scheduled in the first to third lessons in the morning, but the first lesson in the morning cannot be History, the second lesson in the morning cannot be Chinese, the third lesson in the morning cannot be Math, the fourth lesson in the morning cannot be PE, the first lesson in the afternoon cannot be Music, and the second lesson in the afternoon cannot be Geography. How many different schedules are there?
(1) Essentially, it is the Bernoulli-Euler wrong envelope problem when $n=6$, so the number of different ways to arrange the schedule is
$$
\begin{aligned}
u_{5} & =6!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}\right) \\
& =265 \text { (ways). }
\end{aligned}
$$
(2) Essentially, it is the Bernoulli-Euler wrong envelope problem when $n=5$, so the number of different ways to arrange the schedule is
$$
\begin{aligned}
u_{5} & =5!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right) \\
& =44 \text { (ways). }
\end{aligned}
$$
(3) If "Geography is scheduled for the second class in the afternoon," the number of ways to arrange the schedule is $u_5$; if "Geography is not scheduled for the second class in the afternoon," the number of ways to arrange the schedule is $u_6$. Since the restriction "Geography cannot be scheduled for the second class in the afternoon" is removed, the arrangement of Geography can be divided into two scenarios: "Geography is scheduled for the second class in the afternoon" and "Geography is not scheduled for the second class in the afternoon." Therefore, the number of different ways to arrange the schedule is
$$
u_{5}+u_{6}=44+265=309 \text { (ways). }
$$
(4) Essentially, this is based on the third question, removing the restriction "Mathematics cannot be scheduled for the first class in the afternoon," so the arrangement of Mathematics has two scenarios: "Mathematics is scheduled for the first class in the afternoon" and "Mathematics is not scheduled for the first class in the afternoon." Therefore, the number of different ways to arrange the schedule is
$$
\begin{aligned}
& \left(u_{4}+u_{5}\right)+\left(u_{5}+u_{6}\right) \\
= & (9+44)+309 \\
= & 362 \text { (ways). }
\end{aligned}
$$
(5) "Music must be scheduled for the second class in the morning or the fourth class in the morning," which can be divided into two scenarios: "Music is scheduled for the second class in the morning" and "Music is scheduled for the fourth class in the morning." The former, as known from the second question, has $u_5$ different ways to arrange; the latter, considering the restriction "Chinese cannot be scheduled for the fourth class in the morning," the third class can be scheduled for Chinese or not, similar to the third question, this has $\left(u_{4}+u_{5}\right)$ different ways to arrange. Therefore, the number of different ways to arrange the schedule is
$$
\begin{aligned}
& u_{5}+\left(u_{4}+u_{5}\right) \\
= & 44+(9+44) \\
= & 97 \text { (ways). }
\end{aligned}
$$
(6) First, arrange the first four classes, which has $u_4$ ways to arrange; then arrange the two classes in the afternoon, which has 2! ways to arrange. Therefore, by the multiplication principle, the number of different ways to arrange the schedule is
$$
2 \text { ! } \cdot u_{4}=2 \times 9=18 \text { (ways). }
$$
(7) First, arrange the first to third classes in the morning, which has $u_3$ ways to arrange; then arrange the fourth class in the morning and the two classes in the afternoon, which also has $u_3$ ways to arrange. Therefore, by the multiplication principle, the number of different ways to arrange the schedule is
$$
\begin{aligned}
\left(u_{3}\right)^{2} & =\left[3!\left(\frac{1}{2!}-\frac{1}{3!}\right)\right]^{2} \\
& =4(\text { ways }) .
\end{aligned}
$$