There are lot of similar questions available. After going through all, I am still confused with two different answers.

Question : How many ways can three married couples sit around a round table if husband and wife must sit together?

Base on the explanation available in the book , I get the answer as $(3-1)! * 2^3 = 16$. This explanation is also clear to me.

However, after reading further here, the number of ways in which three couples can be arranged around a circular table so that no wife sits next to her husband $= 120 -144 + 72 -16 = 32$ (based on inclusion and exclusion). If so, the number of ways where a wife sits next to her husband should be $5! - 32 = 88$.

I know I am wrong. I am still trying hard to understand where I have gone wrong. Please help.

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    $\begingroup$ The $88$ (which I have not checked) is the number of ways at least one pair of partners are next to each other. But not necessarily all. So there are quite a few more than the $16$ where all pairs of partners are next to each other. $\endgroup$ – André Nicolas Jun 14 '15 at 2:21
  • $\begingroup$ thanks André Nicolas. that means 16 is the answer when every couples sits together , isn't it? $\endgroup$ – Kiran Jun 14 '15 at 2:39
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    $\begingroup$ You are welcome. Yes, $16$ is for all pairs together, $88$ is for at least one pair together. $\endgroup$ – André Nicolas Jun 14 '15 at 2:44
  • $\begingroup$ thanks André Nicolas, I am clear now. how may I close this question now and select your answer? $\endgroup$ – Kiran Jun 14 '15 at 2:45
  • $\begingroup$ To close things up, since the issue may come up for someone else, I wrote an answer. $\endgroup$ – André Nicolas Jun 14 '15 at 3:08

We check your count of the number of ways at least one couple are in adjoining chairs. As you did, we use Inclusion/Exclusion. Call the couples A, B, and C.

The number of ways couple A are next to each other is $(2)(4!)$. For we may assume the fatter of the two members sits in a specific chair. Then the partner has $2$ choices, and for every such choice the rest can arrange themselves in $4!$ ways.

Add together the number of ways for couple A to be together, couple B to be together, couple C to be together. We get $144$.

But we have overcounted the ways in which for example couple A and B are both together. Again we can assume the fatter of the members of A sits in a specific chair. Then there are $2$ choices for where the other member sits, and then $(3)(2)$ ways for couple B to be together, leaving $2$ choices for the others, a total of $24$. Summing over all ways to choose $2$ couples, we get $72$. So our corrected estimate is $144-72$.

But the $144$ counted three times the arrangements in which all the couples are together, and so did the $72$, so we must add back the $16$ ways in which they are all together. The final count is $144-72+16$.

Note that this $88$ counts the number of ways at least one couple are together. It is naturally a much larger number than the $16$ ways in which all couples are together.


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