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a. with no other restrictions 16!/16 = 15! b. every married couple dances together 7! for the couples and 2^8 for alternating husband and wife. 7!*2^8=1296240 c. exactly 3 couples dance together (the wife dances beside the husband). I believe is (2!2^3) (10!) or (3!*2^3) * 9! can somebody help? thanks in advance

gav
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  • That's in $15!$ ways, isn't it? – Allawonder Dec 09 '20 at 16:06
  • a and b look right to me. Isn't part c. $3!(2^3)(10!)$? So you have '11' groups. You have the 3 couples (6 people in one group). And you have the 10 other people. So you have $11!/11=10!$ ways to arrange the 11 groups. But then you have to multiply by $3!(2^3)$ ways to arrange within the group of 6. – Ameet Sharma Dec 09 '20 at 16:19
  • for c I am thinking divide in two cases. – gav Dec 09 '20 at 16:27
  • @gav, I don't understand. Do you disagree with $3!(2^3)(10!)$? – Ameet Sharma Dec 09 '20 at 16:40
  • for c I am thinking first assuming that if they were on a line then: 3!2^6 for the 3 groups and 10! for the rest that is (3!2^6)*10!. Since we have 16 people on the circle we must divide this by sixteen, as if I rotate every result 16 times I have the same arrangement. Am I correct? – gav Dec 09 '20 at 16:41
  • @gav, I think it's easier if you fix the group of six at the beginning of the line. And arrange the other people after the first group. So you get $3!2^3$ for the arrangements within the first group. Then multiply by $10!$ for the arrangements of the people following. Since you've fixed the one group at the beginning of the line, you've already taken the circle rotation into account. – Ameet Sharma Dec 09 '20 at 16:49
  • Can you confirm what c) is asking for exactly? Is it the same 3 couples together in every permutation? – Ameet Sharma Dec 09 '20 at 17:18
  • I think the question may need more clarity. Have deleted my answer for now. – Math Lover Dec 09 '20 at 17:40
  • @MathLover, yeah I assumed some particular set of 3 couples dance together as part of 1 group. But I'm not sure that's the right interpretation. Will wait for OP to clarify. – Ameet Sharma Dec 09 '20 at 17:45
  • clarification for c. 3 particular married couples dance side by side and the rest do not have any restriction. (am,af),(bm,bf), (cm, cf) are together, of course, can be also (af,am),( bf, bm) (cf,cm). The rest can be arranged with no restriction. – gav Dec 09 '20 at 21:55
  • @gav, in that case I think my answer of $3!(2^3)10!$ is the right one. – Ameet Sharma Dec 10 '20 at 00:59
  • thanks but how you take into account the fact that the dance is circular – gav Dec 10 '20 at 18:45
  • @gav, if the dance was in a line it would be $3!(2^3)11!$. We have 11 sets. 10 people + a group of 6 people. Since it is a circular dance we change the $11!$ to $10!$. We can picture this by fixing the set of 6 people. And arranging the 10 other people after them in clockwise order. So we get $10!$ instead of $11!$. And that gives our answer as $3!(2^3)10!$ – Ameet Sharma Dec 11 '20 at 01:16
  • @Ameet, this was my first thought, but then I thought that this may be not correct since the pairs of the married couples can be in any position between the single ones, I cannot consider them as a single group. so I tend to believe it is 3!(2^3)10!/13 (13 entities are 3 groups and 10 people). so 13 places that an entity can rotate and still have the same arrangement. – gav Dec 12 '20 at 20:13
  • oh, so the 6 people are not all together in one group? ok. I was misunderstanding then. In that case it would be $12!(2^3)$ wouldn't it? We have 13 entities. We do $12!$ since they are in a circle. And 3 of the groups (the together-couples) can be arranged 2 ways. So we multiply by $2^3$. – Ameet Sharma Dec 12 '20 at 20:17
  • 3! are the different ways that the 3 pairs can be arranged. 2^3 is the different ways wife and husband can be arranged assuming that wife and husband can alternate that gives a total of 2^3 different arrangements. finally, the 10 singles give 10! different ways. So we have a total of 3!(2^3)10! arrangements. Until this point, I believe I am right. Then, we should divide by 13 to take into account that they are on a cycle. The number 13 I am not sure about. – gav Dec 12 '20 at 23:23
  • ok. But each of the 3 couples can be anywhere right? Just as long as each couple remains together? So we can have 4 regular people, 1 couple, 3 regular people, 1 couple, 2 regular people, 1 couple, 1 regular person. would this configuration be allowed? In that case $3!10!$ does not get all the permutations. We need $13!$ then divide by 13 to get $12!$ – Ameet Sharma Dec 13 '20 at 00:12
  • ok but every pair can have 2 different arrangements (wife-husband) and (husband-wife). that is in total 2^3 different ways for the 3 paiirs. I have no doubt about the 3!(2^3)10! but I am wondering if I must divide by 13 or 16 or something else. – gav Dec 13 '20 at 10:25
  • yes, I was not including the $2^3$ part which we both already agree on. I was just talking about the groups. Anyway, I think the answer is $12!(2^3)$. That's taking $13!/13$ then multiplying by $2^3$. We just have to agree to disagree I guess. – Ameet Sharma Dec 13 '20 at 10:38
  • Ameet, I appreciate your help, but I disagree with 12!. We have 3! for the particular pairs and 2! for alternate and 10! for the rest that makes a total of 3!(2^3)10! and then we must most probably divide by 16. The only part I have a doubt about is about 16. – gav Dec 13 '20 at 20:53

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