How many ways are there for sitting n families around a circular table. Each family is a mother a father and a child.

Condition: The mother and father of each family should be sitting next to each other while children are not allowed to sit next to each other.


  • $\begingroup$ What do you regard as distinct ways? Would you say the answer was 1 for $n=1$? $\endgroup$ – almagest Apr 5 '16 at 13:37
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    $\begingroup$ For n=1 the answer would be 2 as the conditions do not apply here and the number of arrangement of 3 people around a circular table is (3-1)!=2 $\endgroup$ – Zoltán Apr 5 '16 at 13:55

I assume you regard arrangements obtained by rotations and reflections as equivalent.

Start with the location of $M_1$ and choose the direction by taking the next person as $F_1$. Then there are $(n-1)!2^{n-1}$ ways of arranging the other parents and $n!$ ways of fitting the children in the $n$ gaps between them. So total $n!(n-1)!2^{n-1}$.

  • $\begingroup$ Arranging n couples of F's and M's should be (n-1)! multiplied by 2^n as each couple could be sitted as MF or FM. Now we need to arrange the n children and that gives us n! options. To conclude, (n-1)!* 2^n * n!. Could you spot the mistake I am making? $\endgroup$ – Zoltán Apr 5 '16 at 13:51
  • $\begingroup$ need divide for circle rotations $\endgroup$ – miniparser Apr 5 '16 at 13:58
  • $\begingroup$ @miniparser I see. Could you elaborate more please? $\endgroup$ – Zoltán Apr 5 '16 at 14:00
  • $\begingroup$ @Tamir The two questions of definition are: (1) do you regard 1,2,3,4 as the same arrangement as 2,3,4,1 (ie differ only in rotation); (2) do you regard 1, 2, 3, 4 as the same arrangement as 1,4,3,2 (ie going the opposite way round the circle). I assumed the answer was yes in both cases. If your answer to (2) is No, then you need to double the number. $\endgroup$ – almagest Apr 5 '16 at 14:05
  • $\begingroup$ I think divide by 3n $\endgroup$ – miniparser Apr 5 '16 at 14:10


Let us name the mother=$M$,father=$F$ and child=$C$.

As the mother and father sits together,group one family into two parts-$(M,F);(C)$.(How many groups can be formed out of n families if one family breaks up into $2$ groups?).

Now,arrange n number of $(M,F)$ and $(C)$'s around the table by using circular permutation formula that x number of objects can be arranged into $(x-1)!$ ways.

Hope my answer helps!!

  • $\begingroup$ Thought 2: There are four options to sit the first family MFC,FMC,CMF,CFM then for the next n-1 families I have only two options left (for example, MFCCFM won't comply with condition while MFCMFC or MFCFMC will) which is 4*2^n-1. All we have left are the kids that have (n-1)! options to be seated. Final answer: 4*2^n-1*(n-1)! = 2^n+1 * (n-1)! . Is this anyway around? $\endgroup$ – Zoltán Apr 5 '16 at 13:52
  • $\begingroup$ @Tamir Yeah...you are on the right way I think... $\endgroup$ – tatan Apr 5 '16 at 13:56

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