How many possibilities are there to place $k$ persons on a round table with $n$ seats in such a way that there is at least one free seat between each $2$ persons?
The seats are not distinguishable, meaning that rotationally symmetric arrangements are equal.
Does it make sense to say $\frac{k+(n-2k)!}{n-2k} =$ (Persons + Free Seats)! / Free Seats because they are not distinguishable?
I think it's wrong.
Best Answer
First, let's choose where there are people. To do that, we consider the $k$ people as "slots" for people. Between them are "slots" for empty chairs. We want to decide where to place empty chairs. Because there are $k$ people, there are $k$ slots. Place one empty seat in each slot. Now, we want to choose how many additional chairs to place in each slot. This is a stars and bars problem. There are $k$ slots and $n-2k$ remaining seats that need to be placed. So, there are:
$$\dbinom{n-2k+k-1}{k-1} = \dbinom{n-k-1}{k-1}$$
ways to place the remaining seats.
Next, arrange the people. There are $(k-1)!$ circle-arrangements for the people to their chosen slots.
Total:
$$\dbinom{n-k-1}{k-1}(k-1)! = \dfrac{(n-k-1)!}{(n-2k)!}$$