9. A department arranges the duty schedule for four people, Jia, Yi, Bing, and Ding, during the eight-day National Day holiday. It is known that Jia and Yi each work four days, Jia cannot work on the first day and Jia and Yi do not work on the same day; Bing needs to work three days and cannot work consecutively; Ding needs to work five days; it is stipulated that two people must work each day. Then the number of different schemes that meet the conditions is ( ).
(A) 400
(B) 700
(C) 840
(D) 960
9. B.

First, consider A and B. A does not work on the first day, so A can choose any four days from the remaining seven days, thus there are $\mathrm{C}_{7}^{4}$ ways to arrange, and the remaining days are assigned to B, so B has $\mathrm{C}_{4}^{4}$ ways to arrange.
, Suppose C's working days are arranged on
$B_{2}$ days. Then, by the problem, we know $n_{1} 、 n_{2}$ 、
$$
\begin{array}{l}
\leqslant n_{1}<n_{2}<n_{3} \leqslant 8 . \\
2=n_{2}-1, m_{3}=n_{3}-2 . \text { Then }
\end{array}
$$
$$
1 \leqslant m_{1}<m_{2}<m_{3} \leqslant 6 .
$$

Obviously, $n_{1} 、 n_{2} 、 n_{3}$ being non-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, C's working schedule has $\mathrm{C}_{6}^{3}$ different schemes.
D's working schedule is $\mathrm{C}_{5}^{5}$.
By the multiplication principle, the number of different working schedules for A, B, C, and D is $\mathrm{C}_{7}^{3} \mathrm{C}_{4}^{4} \mathrm{C}_{6}^{3} \mathrm{C}_{5}^{5}=700$.