11. There are 11 students who have signed up for the volunteer tour guide activity at the museum. The activity runs from 9 AM to 5 PM, with a public mini-lecture every hour. Each session requires 1 student to provide tour guide services to visitors. To avoid overworking the students, the museum will not schedule the same person for two consecutive sessions, and the first and last sessions must be handled by two different students. How many scheduling arrangements are possible for the museum?
Let the number of scheduling ways for 11 students to explain for $n$ hours be $a_{n}$.
If the last student explaining is different from the first one, then the corresponding number of scheduling ways is $a_{n}$; if the last student explaining is the same as the first one, removing the last session, the corresponding number of scheduling ways is $a_{n-1}$. Thus, $a_{n}+a_{n-1}=11 \times 10^{n-1}(n \geqslant 3)$, obviously $a_{1}=11, a_{2}=11 \times 10=110$.
Since $a_{n}-10^{n}=-\left(a_{n-1}-10^{n-1}\right) \Rightarrow a_{n}-10^{n}=\left(a_{2}-100\right) \cdot(-1)^{n-2} \Rightarrow a_{n}=10^{n}+10 \cdot(-1)^{n}(n \geqslant 2)$, therefore $a_{8}=10^{8}+10=100000010$.