If $k$ people are seated in a random manner in a row containing $n$ seats ($n > k$), what is the probability that the people will occupy $k$ adjacent seats in the row?
If $k$ people are seated in a random manner in a circle containing $n$ chairs ($n > k$), what is the probability that the people will occupy $k$ adjacent chairs in the circle?
For the first question, my solution is $\frac{(n-k+1)!k!}{n!}$. Is this correct? The solution to the second question is similar, but how do I take into account the circle instead of a line?
Best Answer
There are $n-k+1$ possible locations for $k$ people occupying adjacent seats, and there are $\binom{n}k$ possible locations for $k$ people, so the probability in the first question is $$\frac{n-k+1}{\binom{n}k}=\frac{(n-k+1)k!(n-k)!}{n!}=\frac{(n-k+1)!k!}{n!}\;,$$ as you say.
In the second question there are still $\binom{n}k$ possible choices of $k$ seats, but there are now $n$ of them that have the $k$ people in adjacent seats, so the probability is
$$\frac{n}{\binom{n}k}=\frac{nk!(n-k)!}{n!}=\frac{k!(n-k)!}{(n-1)!}\;.$$
(I’m assuming that the seats in the circle are individually identifiable, i.e., that seatings that differ by a rotation are still different seatings.)