Question

Find the number of possibilities in seating arrangements of 3 couples around a circular
table if every woman sits next to her own husband.


This permutation type​

Answer 1

To find the total number of arrangements in which no couple sits together, one must take the total number of arrangements of people and subtract permutations in which one, two, or all three couples are seated next to each other.

Let A1 be the event in which the first couple sit together.

Let A2 be the event in which the second couple sit together.

Let A3 be the event in which the third couple sit together.

There are 6 seats. If we could sit any person anywhere, there would be 6! ways of seating them.

The event Ai can occur in 5! * 2 ways - there are 5 different seat combinations one couple can choose while sitting together, and 4! ways to seat the other four people - 5 * 4! = 5!. For each of these combinations, there are 2 ways to "flip" the couple, hence 5! * 2.

The events A1 ∩ A2, A1 ∩ A3, and A2 ∩ A3 can occur in (5 * 2)(3 * 2)(1 * 2) ways (or 5!) - 5 ways to seat the first couple next to each other, 3 ways to seat the second couple next to each other in the remaining 4 seats, and 2! ways to seat the remaining two people - 5 * 3 * 2!. There are 2 ways to flip each couple, hence (5 * 2) and (3 * 2). This comes together in the form (5 * 2)(3 * 2)(2!), or (5 * 2)(3 * 2)(1 * 2).

The event A1 ∩ A2 ∩ A3 can occur in (5 * 2)(2 * 2)(1 * 2) ways. This is a lot like the aforementioned intersection events, except for one small difference - the second couple can only be seated in 2 different places. This is because if they sit directly across from the first couple such that only one seat is open to either side of them, the third couple is unable to sit together.

The event A1 contains the events A1 ∩ A2 and A1 ∩ A3, both of which contain the event A1 ∩ A2 ∩ A3. The same can be said about A2 and A3, except with their own respective intersections. This must be accounted for. The total is then:

6!−(A1−A1∩A2−A1∩A3+A1∩A2∩A3)−(A2−A1∩A2−A2∩A3+A1∩A2∩A3)−(A3−A1∩A3−A2∩A3+A1∩A2∩A3)−(A1∩A2−A1∩A2∩A3)−(A1∩A3−A1∩A2∩A3)−(A2∩A3−A1∩A2∩A3)−(A1∩A2∩A3)

This is a very exhaustive form of the equation, but it helps me visualize the problem a little bit better - each term surrounded by parentheses represents a section of a Venn Diagram. The simplified form of the equation is as follows:

6!−A1−A2−A3+A1∩A2+A1∩A3+A2∩A3−A1∩A2∩A3

Substitution yields:

6!−3∗(6∗4!∗2)+3∗[(6∗2)(3∗2)(1∗2)]−[(6∗2)(2∗2)(1∗2)]

 

or:

720−3∗288+3∗144−96=192

192/720=.2666666666≈26.6%