What is the probability that, when 3 couples are seated around a circular table, no couple sits together?

Here is the question, straight from the textbook:

If three couples are seated around a circular table, what is the probability that no wife and husband are beside one another?

Here's my take on it.

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 $A_1$ be the event in which the first couple sit together.
Let $A_2$ be the event in which the second couple sit together.
Let $A_3$ 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 $A_i$ 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 $A_1$ $\cap$ $A_2$, $A_1$ $\cap$ $A_3$, and $A_2$ $\cap$ $A_3$ 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 $A_1$ $\cap$ $A_2$ $\cap$ $A_3$ 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 $A_1$ contains the events $A_1$ $\cap$ $A_2$ and $A_1$ $\cap$ $A_3$, both of which contain the event $A_1$ $\cap$ $A_2$ $\cap$ $A_3$. The same can be said about $A_2$ and $A_3$, except with their own respective intersections. This must be accounted for. The total is then:

$$6! - (A_1 - A_1 \cap A_2 - A_1 \cap A_3 + A_1 \cap A_2 \cap A_3) - (A_2 - A_1 \cap A_2 - A_2 \cap A_3 + A_1 \cap A_2 \cap A_3) - (A_3 - A_1 \cap A_3 - A_2 \cap A_3 + A_1 \cap A_2 \cap A_3) - (A_1 \cap A_2 - A_1 \cap A_2 \cap A_3) - (A_1 \cap A_3 - A_1 \cap A_2 \cap A_3) - (A_2 \cap A_3 - A_1 \cap A_2 \cap A_3) - (A_1 \cap A_2 \cap A_3)$$

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:

$$6! - A_1 - A_2 - A_3 + A_1 \cap A_2 + A_1 \cap A_3 + A_2 \cap A_3 - A_1 \cap A_2 \cap A_3$$

Substitution yields:

$$6! - 3 * (5! * 2) + 3 * [(5 * 2)(3 * 2)(1 * 2)] - [(5 * 2)(2 * 2)(1 * 2)]$$

or:

$$720 - 3 * 240 + 3 * 120 - 80 = 280$$

The probability is then found by dividing this number over the total number of arrangements.

$$280/720 = .3888888888 \approx 38.8\%$$

Is this correct?

Nope. First error found: There are 6 ways to sit the first couple together, not 5.

Taking this into account the answer is:

$$6! - 3 * (6 * 4! * 2) + 3 * [(6 * 2)(3 * 2)(1 * 2)] - [(6 * 2)(2 * 2)(1 * 2)]$$

or:

$$720 - 3 * 288 + 3 * 144 - 96 = 192$$

$$192/720 = .2666666666 \approx 26.6\%$$

• answers.yahoo.com/question/index?qid=20121103212212AAzXdUd – zscoder May 7 '14 at 3:07
• @zscoder I saw that and all the answers are completely different. That said, I don't really trust yahoo answers to begin with. – NmdMystery May 7 '14 at 3:08
• Actually I take it back, yeah. Circular desks. Sorry :P Just had finals. – Christopher Liu May 7 '14 at 3:15
• @ChristopherLiu Ah, you're right. – NmdMystery May 7 '14 at 3:16
• @zscoder Actually if I fix the problem ChristopherLiu pointed out, I get the same answer as the top answer to that question, so it's correct and I just didn't understand it. The top answer doesn't really have any explanatory power IMO. – NmdMystery May 7 '14 at 3:34

Label the couples A, B, C so that the six people are A$_{1}$, A$_{2}$, B$_{1}$, B$_{2}$, C$_{1}$ and $C_{2}$.

Now, consider your 6 seats: _ _ _ _ _ _

There are 6 choices for who can sit in the first seat; without loss of generality, say it's A$_{1}$.

There are 4 choices for who can sit in the second seat (anyone else but A$_{2}$); without loss of generality, let's say this is B$_{1}$.

Our table now looks like this: A$_{1}$ B$_{1}$ _ _ _ _

For the 3rd seat, we just can't have B$_{2}$, so we can either have A$_{2}$ or one of the C's; I'll split this into 2 cases.

Case 1: A$_{2}$ sits in the third seat.

A$_{1}$ B$_{1}$ A$_{2}$ _ _ _

Now the fourth seat can't be B$_{2}$, otherwise the C's will sit next to each other in the last two seats. So the fourth seat must be one of the C's; there are 2 choices for this. Then the fifth seat must be B$_{2}$ and the sixth must be the other C.

Thus there are $6*4*1*2*1*1 = 48$ ways for Case 1 to happen.

Case 2: One of the C's sits in the third seat; there are 2 possibilities for this. Let's say it's C$_{1}$

A$_{1}$ B$_{1}$ C$_{1}$ _ _ _

Then there are 2 choices for the fourth seat: A$_{2}$ or B$_{2}$. I'll treat these as subcases.

Case 2.1: A$_{2}$ sits fourth.

A$_{1}$ B$_{1}$ C$_{1}$ A$_{2}$ _ _

For the fifth seat, we have 2 choices, and then 1 for the sixth.

Thus $6*4*2*1*2*1 = 96$ possibilities for this case.

Case 2.2: B$_{2}$ sits fourth.

A$_{1}$ B$_{1}$ C$_{1}$ B$_{2}$ _ _

For the fifth seat, there is only one possibility: A$_{2}$, since otherwise A$_{2}$ would be in the sixth seat, which is adjacent to the first (the table is circular). Of course there is only one possibility for the sixth.

Thus there are $6*4*2*1*1*1 = 48$ ways for Case 2.2 to happen.

There are then $48+96+48 = 192$ possible seating arrangements. The probability of this happening is $\frac{192}{6!} = \frac{192}{720} = \frac{4}{15} \approx 0.267$

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.

So, the equation follows (6-1)!−A1−A2−A3+A1∩A2+A1∩A3+A2∩A3−A1∩A2∩A3

A1+A2+A3= 3 (5-1)! 2! =144

A1∩A2+A1∩A3+A2∩A3= 3 (4-1)! 2! 2! =72

A1∩A2∩A3 = (3-1)! 2! 2! 2! = 16

So,the no couple sits together is

120-144+72-16=32

The sample space is (6-1)!=120

The probabilty is

32/120 = 4/15

• You might want to consider using MathJax for proper math formatting here. – StubbornAtom May 29 '17 at 11:04