there are 4 places available to sit, how many ways can 10 people sit if... there are 4 places available to sit, how many ways can 10 people sit if:

*

*There are 2 people who don't want to sit next to each other

*There are 2 people who want to sit next to each other

I know that this problem uses permutation(order matters) or combination(order doesn't matters) or maybe both, also i'm lost because i don't know if this uses repetition or not and  how to approach the problem.
I know that there are 5040 ways for the people if the 4 places are available, but i don't know where to go from here.
 A: Based on the comments from the question, I'm going to assume the following:
There is one pair, Person A and Person B, such that Person A is selected if and only if Person B is selected, and whenever A and B are selected they sit next to each other.
There is also a third person, Person C, such that whenever Person A is selected, either Person C is not selected or Person C is selected but does not sit adjacent to Person A.
If Person D and Person E are seated in spots 1 and 4, respectively, they are not sitting adjacent to each other, i.e. the table is not a circle.
Both adjacency and handedness matter in the count; i.e. order matters, and flipping the order backwards gives a different solution.
Then let's do some case work!
Case 1: Person A is not selected. Then Person B is also not selected. Any of the other 8 people can be seated with no restrictions. There are $\frac {8!}{4!}=1680$ ways to do this.
Case 2: Person A is selected, but Person C is not selected. We know that Person B must be selected, and then we must choose two other people (out of the 7 remaining) to sit. There are 21 ways to choose the other two people, then 2 ways to choose whether A is on the left or the right of B, then 3 places to choose where A sits (not four, because we already picked what side B was on), and then 2 ways to fill in the other two seats. This
gives a total of $252$ solutions.
Case 3: Person A and Person C are selected. Then Person B is also selected. Pick the last person (7 ways to do this) and call that person Person D. Now, if A is on the edge (fully to the left or to the right), we have 2 ways to pick what side, and then B must be next to A, and then C and D can be in either seat (2 ways to pick). But if A is in one of the two center seats, then it's 2 ways to pick whether it's right-center or left-center, then 2 ways to pick what side B is on, and then C and D only have one option. So that's $7(2\cdot2+2\cdot2)=56$ solutions.
In total, with the assumptions given, there are $1680+252+56=1988$ ways for the people to sit.
