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Find the number of ways in which four boys and four girls can be seated around circular table ?

(a) Find the number of ways in which four boys and four girls can be seated around circular table if the boys and girls are to have alternate seats?

(b) Find the number of ways if they sit alternately and if one boy and one girl are to sit in adjacent seats?

(c) Find the number of ways if they sit alternately and if one boy and one girl must not sit in adjacent seats.

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    $\begingroup$ for (a) I tried (4!/4)*(4!/4)*2 for (b) I tried 7*(3!/3)*(3!/3)*2 and of course (c) equals (b)-(a) $\endgroup$ – Ahmed El-Gamal Mar 13 '15 at 20:59
  • $\begingroup$ are the seats distinct? in other words, if the order is the same, but everyone is shifted to the left, does that count as another way? $\endgroup$ – JLee Mar 13 '15 at 21:09
  • $\begingroup$ No, seats are similar, but boys and girls are distict $\endgroup$ – Ahmed El-Gamal Mar 13 '15 at 21:11
  • $\begingroup$ @AhmedEl-Gamal Welcome to MathSE. For future reference, you are more likely to get a response when you include your work in the statement of the question since that tell us that you have made an effort to solve the problem yourself (which you did) and gives us an idea of why you are encountering difficulties. $\endgroup$ – N. F. Taussig Mar 14 '15 at 0:08
  • $\begingroup$ I agree with you.. $\endgroup$ – Ahmed El-Gamal Mar 14 '15 at 16:41
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Keep in mind that seating arrangements at a circular table are invariant with respect to rotation. Thus, once we seat the first person, all permutations are relative to where that person sits.

(a) Let's sit the girls first. One girl sits down. There are $3!$ ways to seat the other girls. Once the girls sit down, there are $4!$ ways to arrange the boys, giving $$3!4! = 6 \cdot 24 = 144$$ possible seating arrangements in which the boys and girls alternate seats.

Alternate solution: There are $4!$ ways of sitting the girls. However, there are four rotations of the girls that do not change their relative order. Thus, there are $$\frac{4!}{4} = 3!$$ distinguishable ways of sitting the girls. Once they are seated, there are $4!$ ways of sitting the boys. Thus, there are $4!3! = 144$ ways of sitting the girls in which the boys and girls alternate seats.

(b) We seat the boy and girl in adjacent seats first. Once she sits down, he can sit either to her right or to her left, giving two ways of arranging the boy and girl who must sit in adjacent seats. Relative to her, there are $3!$ ways of seating the remaining girls. Once they are seated, there are $3!$ ways of seating the remaining boys, giving $$2! \cdot 3! \cdot 3! = 2 \cdot 6 \cdot 6 = 72$$ seating arrangements in which the boys and girls sit in alternate seats and a particular boy and girl sit in adjacent seats.

Alternate solution: We showed above that we can seat the girls in $3!$ distinguishable ways. There are $2$ ways we can sit the particular boy next to the particular girl (on her right or her left). Once he has been seated, there are $3!$ ways to seat the remaining boys, which yields $$3! \cdot 2 \cdot 3! = 6 \cdot 2 \cdot 6 = 72$$ seating arrangements in which a particular boy sits next to a particular girl.

(c) We subtract the number of ways we can sit a particular boy and girl in adjacent seats when the boys and girls sit in alternate seats from the total number of ways they can sit in alternate seats, which yields $$3!4! - 2!3!3! = 144 - 72 = 72$$ seating arrangements in which the boys and girls sit in alternate seats and a particular boy and girl do not sit in adjacent seats.

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  • $\begingroup$ Once you've calculated your answer for part (a), isn't it clear that in half of those arrangements, the certain girl is one of the two sitting adjacent to the certain boy, and in half of those arrangements, she isn't? Therefore (b) and (c) are both just half of 144. $\endgroup$ – G Tony Jacobs Mar 14 '15 at 5:36
  • $\begingroup$ @GTonyJacobs I think that argument bears some justification. You should post an answer to the question. $\endgroup$ – N. F. Taussig Mar 14 '15 at 11:31
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a) For spot 1, let's say it will be a boy's spot. There are 4 choices. The next spot is a girl's, and there are 4 choices. The next spot is a boy's and there are 3 choices, and so on...

$4*4*3*3*2*2*1*1 = 576$

However, this answer overcounts by a factor of 4, since ABCD = BCDA = CDAB = DABC.

Therefore $576/4 = 144$

b) The first 2 seats are filled by the boy then the girl that must be adjacent to each other. So the remaining seats can be filled in any way, so

$3*3*2*2*1*1 = 36$

but since we can also have the girl then the boy, then we must multiply by 2, so

$2*(3*3*2*2*1*1) = 72$

c) Total number of ways (from part a) - the number of ways that 2 specific people can sit next to each other (from part b)

So, $144 - 72 = 72$

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  • $\begingroup$ the answer in the reference is 3!*4! for (a) and 2*3!*3! for (b) ..for A your answer is much more convincing but for B I think you should multiply it by 2 to order the boy and girl seated next to each other $\endgroup$ – Ahmed El-Gamal Mar 13 '15 at 21:40
  • $\begingroup$ yeah, you are right. I thought of that, but then convinced myself it wasn't necessary. correcting now... $\endgroup$ – JLee Mar 13 '15 at 22:05
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    $\begingroup$ @JLee You overcounted in part (a) since you have not taken invariance under rotation into account. Given a particular seating arrangement of the girls, say Anne, Beth, Carol, and Dalia, the four rotations (Anne, Beth, Carol, Dalia), (Beth, Carol, Dalia, Anne), (Carol, Dalia, Anne, Beth), and (Dalia, Anne, Beth, Carol) leave the girls in the same relative order, so you must divide your answer by $4$. $\endgroup$ – N. F. Taussig Mar 13 '15 at 23:55
  • $\begingroup$ @N.F. Taussig Thanks. I will correct that now. $\endgroup$ – JLee Mar 16 '15 at 14:16

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