(9) A department arranges the duty schedule for four people, Jia, Yi, Bing, and Ding, during the National Day holiday (a total of 8 days off). It is known that Jia and Yi each work 4 days, Jia cannot work on the first day and Jia and Yi do not work on the same day; Bing needs to work 3 days and cannot work consecutively; Ding needs to work 5 days; it is stipulated that two people must work each day. How many different duty schedules meet these conditions?
(A) 400 ways
(B) 700 ways
(C) 840 ways
(D) 960 ways
9 B Method 1 First consider Jia and Yi. Jia is not on duty on the first day, so he can choose any four days from the remaining 7 days, thus there are $\mathrm{C}_{7}^{4}$ ways to arrange, and the remaining can be assigned to Yi as $\mathrm{C}_{4}^{4}$.

Next consider Bing and Ding. Suppose Bing's duty days are arranged on the $n_{1}$th, $n_{2}$th, and $n_{3}$th days of the holiday, then by the problem, $n_{1} 、 n_{2} 、 n_{3}$ are not adjacent, and $1 \leqslant n_{1}<n_{2}<n_{3} \leqslant 8$.

Let $m_{1}=n_{1}, m_{2}=n_{2}-1, m_{3}=n_{3}-2$, then $1 \leqslant m_{1}<m_{2}<$ $m_{3} \leqslant 6$.

Clearly, $n_{1} 、 n_{2} 、 n_{3}$ not being adjacent is equivalent to $m_{1} 、 m_{2} 、 m_{3}$ being distinct, and $\left\{n_{1}, n_{2}, n_{3}\right\}$ and $\left\{m_{1}, m_{2}, m_{3}\right\}$ correspond one-to-one. Therefore, the number of different duty schedules for Bing is $\mathrm{C}_{6}^{3}$. The number of duty schedules for Ding is $\mathrm{C}_{5}^{5}$. By the multiplication principle, the number of different duty schedules for Jia, Yi, Bing, and Ding is $\mathrm{C}_{7}^{3} \cdot \mathrm{C}_{4}^{4} \cdot \mathrm{C}_{6}^{3} \cdot$ $C_{\varepsilon}^{5}=700$.

Method 2 First consider Jia and Yi. Jia is not on duty on the first day, so he can choose any four days from the remaining 7 days, thus there are $\mathrm{C}_{7}^{4}$ ways to arrange, and the remaining can be assigned to Yi as $\mathrm{C}_{4}^{4}$.

Next consider Bing and Ding. If we do not consider the restrictions on Bing, there are $\mathrm{C}_{8}^{3}$ ways to arrange. If Bing works for three consecutive days, there are $\mathrm{C}_{6}^{1}$ ways to arrange. If Bing works for two consecutive days, and it is the first two or the last two days, there are $\mathrm{C}_{2}^{1} \mathrm{C}_{5}^{1}$ ways to arrange. If it is two consecutive days in the middle, there are $\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{1}$ ways to arrange. Therefore, the number of arrangements that do not meet the conditions is $\mathrm{C}_{6}^{1}+\mathrm{C}_{2}^{1} \mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{1}=36$. Therefore, the number of arrangements for Bing and Ding is $\mathrm{C}_{8}^{3}-\left(\mathrm{C}_{6}^{1}+\mathrm{C}_{2}^{1} \mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{1}\right)=56-36=20$.

By the multiplication principle, the number of different duty schedules for Jia, Yi, Bing, and Ding is $\mathrm{C}_{7}^{3} \cdot \mathrm{C}_{4}^{4} \cdot$ $20=700$.