Circular Permutations Question

Note that with 7 people and 5 seats, you can make only 5 people sit. So another way of going about it is you first select the 5 people who are going to sit in 7C5 ways. Now you simply have 5 people and 5 seats around a circle. You can make them sit in 4! ways (we know the (n-1)! formula)
So you get 7C5*(4!) = (7*6/2)*4! = 504
If number of people are more than the seats, the problem can simply be brought down to equal number of people and seats by selecting the people who will sit.

If the number of seats are more, you can increase the number of people by bringing in Mr V (a vacant spot). If you have more than one vacant spots, all Mr Vs are considered identical.
Check out the concept of Mr V (the vacant spot) in linear arrangements here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/10 ... ts-part-i/

Or you can use the same basic counting principle to arrive at the answer in this case too.
Say you have 4 people and 5 seats around a circle. Every person will sit. So start with the first person. He has 1 way of sitting around a circular table because all seats are identical.
Second person has 4 options. (since all seats are distinct now)
Third person has 3 options.
Fourth person has 2 options.
So all 4 can sit in 1*4*3*2 = 24 ways

Now you can extend this logic with any number of seats.
Say you have 4 people and 8 seats around a circle. Again, every person must sit.
Start with the first person who has 1 way since all seats are identical.
The second person can sit in 7 ways since the rest of the 7 seats are distinct now.
The third person in 6 ways.
The fourth person in 5 ways.
So all 4 can sit in 1*7*6*5 = 210 ways.