How many ways if three couples are to be seated in a row and must be beside each other?

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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:

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

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