Combinations of students around 2 circular tables I have a question:

Eleven students go to lunch. There are two circular tables in the dining hall, once can seat $7$ people, the other can hold $4$. In how many ways can the seats be arranged.

I have read this question multiple times to try to understand how to solve How many ways can $p+q$ people sit around $2$ circular tables - of sizes $p,q$?
I get that table 1 will have $p=6!$ permutations of students and table 2 will have $q=3!$ permutations of students and then there needs to be a permutation of how the students are arranged around the two tables. But what I don't understand is how in the formula $${+\choose p}(−1)!(−1)!$$ it is $P(11,7)$? Shouldn't it be $P(11, 4)$ because it is that there are $11$ options of people seated at table 1 and then $4$ options of people remaining to seat at table 2?
And is the answer then $P(11,7)\cdot6!\cdot3!$. The answer in the book is $P(11,7)\cdot3!/7$ which is different from the one I get from the above formula.
 A: 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 $${11\choose 7}\cdot 6!\cdot 3!$$
A: If there are $n$ students, where $k$ of them are to be placed at one specific table and the other $n-k$ have to be placed at a second table, then for each placement, you can independently rotate the students at the first table in $k$ ways and the students at the second table in $n-k$ ways, and this way create $k(n-k)$ different ratotions, without changing the circualar (relative) order. Now, think of the chairs and students to be numbered. Rotating each of the $x$ arrangements will create $k(n-k)$ different permutations (when looking at the numbering), and each permutation can be obtained this way, so: $$x \cdot k \cdot (n-k)=n! \text{ and therefore } x=\frac{n!}{k \cdot (n-k)}$$
This formula is symmetric in $k$ and $n-k$, so it doesen't matter which of the two tables can take 7 students.
Now, with $n=p+q$ and $k=p$, we get $$x=\frac{n!}{k \cdot (n-k)} = \frac{(p+q)!}{p \cdot q}$$ And if you really wish to do so: 
$$
\frac{(p+q)!}{p \cdot q} = \frac{(p+q)!}{p \cdot q} \cdot \frac{(p-1)!}{(p-1)!} \cdot \frac{(q-1)!}{(q-1)!} = \frac{(p+q)!}{p! \cdot q!}(p-1)!(q-1)! = \binom{p+q}{p}(p-1)!(q-1)!
$$
