Let $F(n)$ be the number of ways of seating $n$ people. We could start by carefully listing the ways, for a few small values of $n$. The listing should be almost explicit. We find that $F(1)=1$, $F(2)=2$, $F(3)=6$, and (perhaps) that $F(4)=24$. Despite the small amount of evidence, the conjecture $F(n)=n!$ is tempting.
We prove the conjecture, in principle by induction. Suppose that we know that for a specific $k$, $F(k)=k!$. Now George, the $(k+1)$-th person, comes along, late as usual. For every possible seating of the $k$ people, we can place George at the table of one of the $k$ people, and immediately to the right of that person ($k$ choices), or we can place George at a table by himself. Thus
$$F(k+1)=kF(k)+F(k)=(k+1)F(k)=(k+1)!.$$
Since $F(1)=1$, we conclude that $F(n)=n!$ for all $n$.
Another way: We can use fancier language. For example, note that every permutation of the set $\{1,2,\dots,n\}$ can be expressed uniquely as a product of disjoint cycles. We explicitly include any $1$-cycles. The order in which the product is taken does not matter, since the cycles are disjoint.
Every product of disjoint cycles corresponds to a unique circular seating, and vice-versa. (The people in a cycle determine a table, and their cyclic order determines the order of seating at that table.) This gives an explicit bijection between the permutations of $\{1,2,\dots,n\}$ and the seatings. Thus the number of seatings is equal to $n!$, the number of permutations.
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.
Best Answer
You probably know that $n$ people can be arranged on $(n-1)!$ around the table.
We first choose among $11$ people $7$ people for the first table and the rest go to the second table, that we can do on $${11\choose 7}$$ ways. Now we have to arrange choosen people around tables. For first one we have $6!$ ways and for second one $3!$ ways an thus $${11\choose 7}\cdot 6!\cdot 3!$$