# Dinner party permutations

I was at dinner and we started discussing this problem:

4 people sitting in chairs at dinner, and various people leave to go to the bathroom.

How many total unique permutations are there of people sitting at the table if a unique permutation also depends on which chair you are sitting in (so a unique permutation depends on who is there, and which chair they are sitting in)

I'm not sure whether it would just be 2^4 (the number of combinations of people, including/excluding each of them) times 4! (the rotations in chairs) but then we were trying to figure out if this is over counting or under counting.

## 1 Answer

The formula is given by : $$\begin{gather} \sum_{i=1}^4 {{4}\choose{i}}^2 i! = 208 \end{gather}$$

Explantations: For every possible number i of people around the table, there is $${{4}\choose{i}}$$ ways to choose which people are around the table, $${{4}\choose{i}}$$ to choose the used chairs and $$i!$$ ways to permute the people with the chairs.

• Nice answer. However, the boldface word should be Explanation. Jul 31, 2021 at 9:50