Sitting around a round table in circular permuations If $3$ people are selected at random from $15$ people sitting at a round table, then what is the probability that no two of them are adjacent to each other?

This can be generalized to x people in y seats. As Andre noted, you can use symmetry, but I will assume seat 0 is open.
Let's have "O" refer to an open seat and "F" refer to a filled seat. Then every table combination can be expressed as a combination of (O) and (F) strings, and every legal table combination with seat 0 open can be expressed by a combination of (O) and (OF) strings.
Any seating combination, legal or not, is made up of x (F) and y-x (O) strings. But since it must start with an (O), that leaves y-x-1 (O) and x (F) in the last y-1 slots, so you have ${ y-1 \choose x}$ total combinations.
Any legal seating combination will contain x (OF) and y-2x (O) strings. That means you have ${ y-x \choose x }$ legal combinations.
That makes ${ \frac{y-x \choose x}{y-1 \choose x}}$ our more general answer. I don't think the combinations can be reduced to something simpler, though
${ \frac{(y-2x+1) * ... * (y-x-1)}{(y-x-1) * ... * (y-1)} }$ might help to visualize this. This assumes x > 1, which is okay, since x=0 or 1 is trivial.
