A table for six?

Probability Level 3

A group of three couples sit down for dinner at the table below:

For fun, they decide that no couple can sit across from or next to each other.

In how many ways can they be seated?

Image credit : bayoulog.com

The answer is 96.

1 solution

Geoff Pilling
Sep 20, 2018

There are only two orientations for the three pairs (where one couple either of the opposite corners). And for each of these orientations there are three ways to permute the couples, and for each couple there are two ways they can sit for a given permutation (i.e. they can trade places).

This gives, $2 \cdot 3 \cdot 2 \cdot 2 \cdot 2 = \boxed{96}$ possibilities.

Again, sorry for the misreporting, (same 3! = 3 issue... ugh). This gets quite a bit trickier for $n \gt 3$ couples, as the scenario where couples sit apart but on the same side comes into play. I suspect you're already working on posting the $n = 4$ version, so I'm trying to solve it in anticipation. :)

Brian Charlesworth - 2 years, 8 months ago

Hahaha... Guess I'm pretty predictable... ;-). Actually recently I've been racking my brain over whether there is a general solution for this problem for a triangle with $n$ rows.

Geoff Pilling - 2 years, 8 months ago

Arghh... that was a tough one! I could only find 49; I should have been more patient in finding the rest. I can't even fathom what a general solution might be ....

Brian Charlesworth - 2 years, 8 months ago

@Brian Charlesworth – I was wondering if perhaps some modification of the hook length formula might be able to be used here... 🤔

Geoff Pilling - 2 years, 8 months ago

A table for six?

The answer is 96.

1 solution

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