How many ways can $n$ people sit around two tables? 
There are two different tables, the first have $p$ chairs and the second have $n-p$. How many ways the $n$ people can sit on those tables?

I have two possible answers: $n!$ or ${n \choose p}$.
But I'm not sure if one of those answers is right because my friends told me they are both are wrong.
Any suggestions please, thanks!
 A: Your answer $\ \boldsymbol{n \choose p}\,$ is correct if you don’t account for the sitting order/arrangements. 
Now let's take into consideration the exact way people may be seated around the two tables. For example, if you have 5 persons at one of the tables you can place them in exactly 5! ways. This means your answer will be 
$$\ {n \choose p}\,p!\,(n-p)! = n!$$  if sitting arrangements at those two tables matter. So again your answer is correct here.
SOLUTION TWO (most interesting)
Order matters but we are not ridiculous about it. So we want to account for sitting arrangements in the following fashion. We say that we can place two people around one table in only one possible way – the guys are opposite each other. Then if we have 3 persons at a table, the number of sitting arrangements is still only one arrangement (person 1 will be always next to the other two guys). If we have 4 persons, we can squeeze the fourth person in 3 different places, which will make the number of arrangements to be 3. That matters, of course. Some guy may want to sit next to two specific persons – a reasonable thing to account for. (That's not why we account for it; we just want to be algebraically precise here) Now if we squeeze a fifth person among 4 persons, the number of arrangements will be $4\cdot3$. We can continue like that $\;4\cdot3;\quad 5\cdot4\cdot3;\quad  6\cdot5\cdot4\cdot3; \; .....\; (n-1)(n-2)\cdot...\cdot3 = (n-1)!/2$
$(n-1)!/2\;$ is gonna be  $\;(p-1)!/2\;$  sitting arrangements for the table with $\;p\;$ chairs ($p ≥ 3$). 
And it’s gonna be $\;(n-p-1)!/2\;$ sitting arrangements for the table with $\;n-p\;$ chairs ($n-p-1 ≥ 3$).
Now we have to multiply all possible combinations:
$$ {n \choose p}\, \frac{(p-1)!}{2} \,\frac{(n-p-1)!}{2} = \frac {n!}{(n-p)!\,p!} \,\frac{(p-1)!\, (n-p-1)!}{4} = \frac{n!}{4\,np} = \frac{(n-1)!}{4p} $$
where $p ≥ 3$ and $n-p-1 ≥ 3$ (otherwise it is trivial but the whole formula no longer holds)
Hope my reasoning is clear. Nice problem.
So, your third answer is $\boldsymbol{\;(n-1)!/4p}$ 
and that’s the most reasonable one but I don’t consider your other answers to be incorrect. I hate it when people drag in ambiguity and then it turns out that you are wrong because your answer is supposedly not the best fit.
Choice with $n!$ is the best fit in my opinion. And the people should sit at the tables not on the tables :)
PS: Brrr, my first use of MathJax. Nice implementation.
A: Assume that different permutations of the same $p$ people at the same table are counted as "different ways" to sit people; assume also that one table has $p$ chairs while the other table has $n-p$ chairs. Then the total number of ways is
$$
{n\choose p}\cdot p! \cdot (n-p)! \,=\, n!
$$
where the ${n \choose p}$ coefficient gives the number of ways to select $p$ people out of $n$, while $p!$ and $(n-p)!$ account for different permutations of people at their respective table.
