The prompt is to find the ways there are to sit n people on k chairs standing in a row so that no 2 people sit next to each other.
Pigeon hole theorem is what I am using to attempt this problem. since there are k chairs, and n people. we must have at least $k-n+1$ chairs.
Assuming people to be X and chairs to be O, it can be visualized as XOXOXOX so at least n-1 unoccupied chairs are needed to occupy n people in a way that they aren't next to each other or vice versa.
So now we have $k-n+1$ holes and n pigeons to occupy these places, giving us ${k-n+1} \choose{n}$$n!$ ways to do this right since all people are unique.
Let's say we have 9 chairs and 4 people, we have $6\choose4$$4! = 450$ ways to do so.
Is this the correct way to approach a problem like this?
