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In a restaurant there are four identical circular tables. each tables have 4 chairs. Two families of 7 members and 8 members come to the restaurant. Then find the number of ways these members can be seated (if members of two families do not sit on the same table).

(1)$\frac{(7!)^2}{4}$

(2)$\frac{(8!)^2}{16}$

(3)$\frac{(7!)^2}{16}$

(4)${(8!)^2}$

My approach selection of members into four groups $^8C_4*^7C_4$ .This forms numbers of ways of four distinct groups. Now four group are selected . Lets treat the four group as 4 distinct balls and number of tables as 4 identical balls as the word identical is clearly mentioned but I am not able to solve it

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there's a catch when you divide 8 people in 2 groups of four then you have to divide ${8 \choose 4}$ by 2 (let's say you choose 1,2,3,4 but that's identical with choosing 5,6,7,8).

Now if each table is identical , it does not matter where family members go to, we will say the ways to allocate the tables to 4 groups is '1'.

The ways of arranging 3 groups of 4 people is $(3!)^3$(ways to arrange 4 people on a circle is 3!).

As for the other group with 3 people can be treated 3 people and 1 hole (it matters which 2 persons are around hole),thus making it the case of arranging 4 objects on a circle which again is 3!.

our answer would be $\frac{7!8!3!3!3!3!}{4!4!4!3!2!}=\frac{7!7!\times8}{2^7}=\frac{(7!)^2}{16}$.

sorry if it's a little not understandable.

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