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75 people will sit around a large circular table:

  • 20 members will have a spouse with them
  • 15 members will come alone
  • 5 members will have a spouse and two children

How many different seating arrangements will there be if every family group must sit next to each other at the same large circular table?

My thoughts: We can treat each family group as a person to obtain the following total number of ways: $$ (3-1)!\cdot P(15,15)\cdot P(2,2)^{20}\cdot P(4,4)^5\approx 2.18\times10^{25} $$ But I have a solution that says $$ P(39,39)\cdot P(2,2)^{20}\cdot P(4,4)^{5}\approx 1.70\times 10^{59} $$ Can someone tell me where I'm going wrong and how the solution is right?

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There are $40$ groups, hence, since the table is circular, there are $39!$ ways of placing the groups.

Each of the $20$ groups with two people can be permuted in $2!$ ways.

Each of the $5$ groups with $4$ people can be permuted in $4!$ ways.

Hence the total number of arrangements is $$(39!)(2!)^{20}(4!)^5$$

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  • $\begingroup$ Would my original thought be correct if the condition were that each set of members had to be aligned next to each other in some way? That is, the individuals had to be together, the couples, and then the families... $\endgroup$ – rudy Jun 7 '17 at 15:13
  • $\begingroup$ Yes, but that's a different question. $\endgroup$ – quasi Jun 7 '17 at 15:17

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