# Seating families around a circular table

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?

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 $$(39!)(2!)^{20}(4!)^5$$