Seating of two families of 7 members and 8 members

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

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).
The ways of arranging 3 groups of 4 people is $$(3!)^3$$(ways to arrange 4 people on a circle is 3!).
our answer would be $$\frac{7!8!3!3!3!3!}{4!4!4!3!2!}=\frac{7!7!\times8}{2^7}=\frac{(7!)^2}{16}$$.