How many ways can x and y people be arranged in z seats arranged in a circle? If z > x + y and y arrive after x

Question

For the question

How many ways can x and y people be arranged in z seats if z > x + y and y people arrive after x?
I got (zCx)*((z-x)Cy) if we only want to count combinations, but (zCx)x!((z-x)Cy)y! if we want permutations. The reason being is we can think of it as counting separate experiments. One being x people choosing z seats then y people choosing z - x seats because x of the z seats are taken.

How many ways can they be seated in a circle?:
This is the one I'm not sure of cause circular permutation on wolfram says the solution is (n-1)! for arranging n distinct objects. I'm not really sure how to apply this to the solution above. If my solution to the above is incorrect do let me know as well.

Answer 1

Note that the condition of $y$ people choosing seats after $x$ is actually irrelevant; all possible seatings are equally likely regardless of this condition. So we wish to find the number of ways to seat $x+y$ people at a table with $z$ seats. Since rotations are identical, we find the number of seatings of people $2$ to $x+y-1$ relative to the first person, yielding a total $^{z-1} P_{x+y-1}$.

Answer 2

Apparently the answer was (zCx)(zCy)/z. Working under the assumption that the people in x and y are indistinguishable like the chairs. Then there are z possible rotations to get rid of.