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.
Best Answer
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.