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?

If no-one can sit next to someone else, then three seats are blocked every time someone sits down.  However, some of these may have been blocked already, by previous sitters.  So, the first person to sit down has $15$ choices, and leaves twelve choices.  The next person to sit down has $12$ choices; in two of these cases, he leaves ten choices; in the other ten cases, he leaves nine.  The final person to sit down has either $10$ or $9$ choices, depending on what the second person did.  The result is
$$
N=15\cdot(2\cdot10+10\cdot9)=1650
$$
ways for three (distinguishable) people to sit down.  The probability, then, is
$$
p=\frac{N}{15\cdot14\cdot13}=\frac{1650}{2730}=\frac{55}{91}.
$$
