In how many ways can three couples be seated in a round table

Answer

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Hint: Here, first we have to take the couples as A, B and C. First we have to calculate the number of possible ways that the couples are seated next to each other. Here, each couple has to be arranged among themselves in 2! ways. Each couple can be seated next to each other in 3! ways. With this data we have to calculate the total possible seating arrangements. Next, to find the number of ways that all ladies sit together, we have to arrange the ladies among themselves which is 3! ways and then, consider 3 men and 3 ladies as 4 units where 3 ladies are tied together which will be 4! Ways.Complete Step-by-Step solution:Here, we are given that the married couples are to be seated next to each other in a cinema hall.Now, we have to find the number of possible ways they can be seated and also the number of ways of their seating if all ladies sit together.First let us consider the three couples as A, B and C which each couple is taken as single unit as in figure:

Then the number of possible ways in which the spouses can be seated next to each other is 3! ways.We also know that the couples can interchange their position and sit. Then the number of possible ways that the couple can be seated is 2! ways.Since, there are three couples then each couple can be seated in 2! ways.Therefore, the total number of possible ways that the couples sit next to each other is, n where:$n=3!\times 2!\times 2!\times 2!$We know that:$\begin{align}  & 2!=1\times 2=2 \\  & 3!=1\times 2\times 3=6 \\ \end{align}$Now, by substituting these values we obtain:$\begin{align}  & n=6\times 2\times 2\times 2 \\  & n=48 \\ \end{align}$Therefore, we can say that the spouses can be seated next to each other in 48 possible ways.Next, we have to find the number of ways of seating if all ladies sit together.So, here we will have three ladies sitting together.The three ladies among themselves are arranged in 3! ways.Hence, we will have 4 units, 3 men and 3 ladies tied together. These 4 units can be arranged in 4! ways.Therefore, the total number of possible ways where the ladies sit together = $4!\times 3!$We have:$\begin{align}  & 4!=1\times 2\times 3\times 4=24 \\  & 3!=1\times 2\times 3=6 \\ \end{align}$Hence, the total number of possible ways where the ladies sit together = $24\times 6=144$Therefore, we can say that there are 144 possible seating arrangements if all the ladies sit together.Note: Here, we have to take the couples as a single unit. While arranging we also have to arrange among themselves which will be 2! ways. While arranging the ladies sitting together we will have 4 units, 3 men and 3 ladies tied 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:

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

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