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Three couples are to be seated on one side of a table for lunch. One of the couples, Edward and Bella, are quarreling and would like to be seated apart. How many arrangements are possible if each of the other two couples would like to be seated together?

Can someone please explain how the answer is 48.

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    $\begingroup$ You surprised that $48$ is the answer. Did you get something different? $\endgroup$ – Theo Bendit Feb 23 at 6:45
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    $\begingroup$ Hint: How many ways are there to seat the quarrelers such that one or two couples are between them? How many wyas are there to switch people withon the couples (quarrelling or harmonious)? $\endgroup$ – Toffomat Feb 23 at 6:55
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First seat Edward & Bella. There are two of them and exactly three places (relative to the other couples who sit together.

$$ ? == ? == ?$$

So there are $3 \cdot 2 = 6$ ways of seating Bella & Edward before we seat the other two couples.

Now seat the other two couples. There are two places for the couples, and each couple can switch places with each other, hence there are $6 \cdot 2 \cdot (2 \cdot 2) = 48 $ places.

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    $\begingroup$ thank you. that was helpful $\endgroup$ – bonny Feb 23 at 7:22
  • $\begingroup$ I am sorry I didn't get the last 6*2*2*2. Since there are 2 places and 2 couples , it should be 6*2 and then since they can exchange among themselves it should be 6*2*2. I didn't get how the last 2 came into picture. Can you explain? $\endgroup$ – Vinanth S Bharadwaj Feb 23 at 7:43
  • $\begingroup$ @VinanthSBharadwaj There are two places to put two couples. That is one of the $2$s. Now focus on one of the couples, they can swap places, hence another $2$. Finally the other couple, which gives the last $2$. $\endgroup$ – copper.hat Feb 23 at 19:22
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Well, even I am surprised that the answer given is $48$,
because simply seating couples together and switching husband/wife in $3!*2^3 = 48$ ways !

To keep the quarrelsome couple apart, first seat the two harmonious couples in $2!*2^2 = 8$ ways, then push in two chairs for the quarrelsome couple at two of the uparrows in the diagram below

$\uparrow \fbox{HW}\uparrow \fbox{WH}\uparrow$

so the final answer is $8\times 3\times 2 = 48$

which is the same if they weren't quarrelsome !

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