# Probability people will occupy $k$ adjacent chairs?

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?

-

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
$$\frac{n}{\binom{n}k}=\frac{nk!(n-k)!}{n!}=\frac{k!(n-k)!}{(n-1)!}\;.$$
Would you elaborate why there are $n$ choices rather than $n-k+1$? –  idealistikz Dec 22 '12 at 9:17
@idealistikz: Number the seats $0$ to $n-1$ clockwise. For each $m\in\{0,\dots,n-1\}$ there is a block of $k$ consecutive seats numbered $m,m+1,\dots,m+k-1$, where the addition is performed modulo $n$. And each block of $k$ consecutive seats is formed in this way. Less technically, if we stand in the middle of the circle and turn clockwise, the first person in the block of $k$ can be sitting in any one of the $n$ seats. –  Brian M. Scott Dec 22 '12 at 9:24