Keep in mind that seating arrangements at a circular table are invariant with respect to rotation. Thus, once we seat the first person, all permutations are relative to where that person sits.
(a) Let's sit the girls first. One girl sits down. There are $3!$ ways to seat the other girls. Once the girls sit down, there are $4!$ ways to arrange the boys, giving
$$3!4! = 6 \cdot 24 = 144$$
possible seating arrangements in which the boys and girls alternate seats.
Alternate solution: There are $4!$ ways of sitting the girls. However, there are four rotations of the girls that do not change their relative order. Thus, there are
$$\frac{4!}{4} = 3!$$
distinguishable ways of sitting the girls. Once they are seated, there are $4!$ ways of sitting the boys. Thus, there are $4!3! = 144$ ways of sitting the girls in which the boys and girls alternate seats.
(b) We seat the boy and girl in adjacent seats first. Once she sits down, he can sit either to her right or to her left, giving two ways of arranging the boy and girl who must sit in adjacent seats. Relative to her, there are $3!$ ways of seating the remaining girls. Once they are seated, there are $3!$ ways of seating the remaining boys, giving
$$2! \cdot 3! \cdot 3! = 2 \cdot 6 \cdot 6 = 72$$
seating arrangements in which the boys and girls sit in alternate seats and a particular boy and girl sit in adjacent seats.
Alternate solution: We showed above that we can seat the girls in $3!$ distinguishable ways. There are $2$ ways we can sit the particular boy next to the particular girl (on her right or her left). Once he has been seated, there are $3!$ ways to seat the remaining boys, which yields
$$3! \cdot 2 \cdot 3! = 6 \cdot 2 \cdot 6 = 72$$
seating arrangements in which a particular boy sits next to a particular girl.
(c) We subtract the number of ways we can sit a particular boy and girl in adjacent seats when the boys and girls sit in alternate seats from the total number of ways they can sit in alternate seats, which yields
$$3!4! - 2!3!3! = 144 - 72 = 72$$
seating arrangements in which the boys and girls sit in alternate seats and a particular boy and girl do not sit in adjacent seats.
We consider the 3 cases mentioned above:
1) If they are split 4/1/1, there are $\dbinom{6}{4}$ ways to select the 4 people sitting together, and $3!$ ways to arrange them at their table, so this gives $\dbinom{6}{4}(3!)=15(6)=90$ possibilities.
2) If they are split 3/2/1, there are $\dbinom{6}{3}$ ways to select the 3 people sitting together, $2!$ ways to arrange them at their table, and $\dbinom{3}{2}$ ways to select the 2 people sitting together, so this gives $\hspace{.2 in}\dbinom{6}{3}\cdot2\cdot\dbinom{3}{2}=20(2)(3)=120$ possibilities.
3) If they are split 2/2/2, then there are $\displaystyle\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}$ or $\;5\cdot3\cdot1=15$ ways to pair up the 6 people.
Therefore there are a total of $90+120+15=$225 possible ways to do this.
This is s(6,3), a Stirling number of the first kind.
Best Answer
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.