Solution: 2014 Winter Final - 8

Author: Michiel Smid

Question

Consider a group of $n$ people, let $k$ be an integer with $1 \leq k \leq n$, and consider a circular table with $k$ chairs around it. We select $k$ people and seat them around this table. How many different seating arrangements are there? (Two seating arrangements $A$ and $B$ are the same if for each person, the clockwise neighbor in $A$ is the same as the clockwise neighbor in $B$, and the counterclockwise neighbor in $A$ is the same as the counterclockwise neighbor in $B$.)

(a)

$\frac{n!}{k(n-k)!}$

(b)

$\frac{n!}{k!(n-k)!}$

(c)

$\frac{2 \cdot n!}{(n-k)!}$

(d)

$\frac{n!}{(n-k)!}$

Solution

We need to determine the number of different seating arrangements.

We choose $ k $ people from $ n $ people: $ \binom{n}{k} $
We also need to find the number of circular arrangements/permutations
It's important to identify that in a table of 3 people, $ (1,2,3), (2,3,1), and (3,1,2) $ are the same arrangement because 2 always has 1 on their left and 3 on their right
Therefore, we need to divide the number of linear arrangements by $ k $ to account for the rotations
$ (k-1)! $

$ \binom{n}{k} \cdot (k-1)! $

$ \frac{n!}{k!(n-k)!} \cdot (k-1)! $

$ \frac{n!}{k!(n-k)!} \cdot \frac{(k-1)!}{1} $

$ \frac{n!}{k(k-1)!(n-k)!} \cdot \frac{(k-1)!}{1} $

$ \frac{n!}{k(n-k)!} $