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In how many ways can 2 adults, 2 girls, and 2 boys be seated around a circular table if the adults are to sit together and the boys and girls are to alternate?

My answer was that there's 8 ways - the adults can swap, while the boys and girls each have 2! ways of sitting. Is this correct?

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    $\begingroup$ You could have AABGBG or AAGBGB. $\endgroup$ – Lord Shark the Unknown Sep 14 '18 at 6:34
  • $\begingroup$ People are all distinct though. $\endgroup$ – globe1004 Sep 14 '18 at 6:35
  • $\begingroup$ Do you consider a mirror image distinct? Everyone still sits next to their same partners, just some have swapped left/right partners. E.g. to explain: i.imgur.com/iQUbY8a.png $\endgroup$ – orlp Sep 14 '18 at 6:38
  • $\begingroup$ If you are talking about a rotation of the same seating arrangement, then no, that's not distinct. $\endgroup$ – globe1004 Sep 14 '18 at 6:42
  • $\begingroup$ @globe1004 I'm not talking about rotation. See the attached image. You can't get one scenario from the other by rotating. Yet still everyone has the same partners in both cases. $\endgroup$ – orlp Sep 14 '18 at 6:42
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The parents can swap, the boys can swap and the girls can swap. That's $8$. However, in addition, you can also have the boys swap with the girls, which makes the answer $16$.

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  • $\begingroup$ "However, in addition, you can also have the boys swap with the girls, which makes the answer 16"...Wouldn't there be double-ups? I thought that'd be considered as a rotation of the same seating arrangement. $\endgroup$ – globe1004 Sep 14 '18 at 6:38
  • $\begingroup$ No. There are $8$ possible seatings where the kid immediately to the right of the parents is a girl, and $8$ where the kid immediately to the right of the parents is a boy. These two cases have no overlap. It's not a rotation because the seating of the parents locks down which rotation you have; any rotation would make the parents sit somewhere else. It is, however, a mirroring, but that ought to count as distinct, circular table or not. $\endgroup$ – Arthur Sep 14 '18 at 6:40
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Let the two adults be a mom and a dad. Let's view the problem from the perspective of the mom through four independent observations:

  1. Her husband can be either to her left or the right, giving a factor of $2$.

  2. Wherever her husband does not sit, either a daughter or a son can sit, giving a factor of $2$.

  3. The child next to her is either the first or the second child of that sex, giving a factor of $2$.

  4. The child of the opposite sex that sits one space away from her can be either the first or the second, giving a factor of $2$.

These four independent observations fully describe the state of the table for $2^4 = 16$ possibilities.

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We can have 2 patterns: $AAbgbg$ or $AAgbgb$.

In each of these we can permutate inside groups of adults, boys, and girls, each in $2!$ ways. Therefore we have $$2\cdot 2!\cdot 2! \cdot 2! = 16$$ ways of arranging the seating.

Anyway, if there are adults with different sexes and we don't want Males sit next to Females, we still have 2 patterns: $MFbgbg$ and $FMgbgb$, but the adults can't permutate, therefore in this situation we have $$2\cdot 2!\cdot 2! = 8$$ ways.

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My answer is 2! x 2 x 2 x 2 = 16.
Reasoning:
1. consider swapping parents
2. then choose boys or girls to group with parents
3. with that group, permutate with table (i.e (3-1)!)
4. then there will be 2 slots to fit between BGB (I assume that boy is group with parents in step 2 & 3)

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In case of round table we have two tasks to counter excess to linear problem:

1) Fixing the end. 2)Countering the direction.

In this case we fix the position by two 2 parents who can permute in two ways. Then going in clockwise sense we can fix any one of the 4 kids in 4 ways then we shall have only one option for rest of positions keeping in mind the restriction.

We can repete same process in counterclockwise position.

Hence we shall have $2(1×2×4×1×1×1)$ arrangements

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  • $\begingroup$ We only need to consider the clockwise sense. However, once we choose the order of the parents and the child who is next to the parents as we proceed clockwise, we must also choose which child of the opposite sex is next to that child. After that, all the positions are completely determined. $\endgroup$ – N. F. Taussig Sep 14 '18 at 10:31

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