Hard permutations and combinations problem

Question

The question says

A minibus has 11 passenger seats. There are six seats in a row on the sunny side and five seats in a row on the shady side, as shown in the following diagram.

Find how many ways eight passengers can be arranged in these seats if:

a) there are no restrictions

b) one particular passenger refuses to sit on the sunny side

c) two particular passengers refuse to sit in seats that are either next to each other or one directly in front of the other.

Here is what I did for (a) and (b):

a) $11P8=6652800$

b) $5C1\times 10P7=3024000$

I am now stuck on (c). The answer is given as $4959360$.

Answer 1

The answer is $11P8-2*(4+5+5)*9P6=4959360.$ Can you see why?