if there are 4 different tables and 18 people, how many ways can the people be seated to have at least 4 in each table? For example with 15 people at 3 different tables each seating 5 people, is the number of combinations of seating equal to:
$$\binom{15}5\binom{10}5(4!)^3$$
 A: There are many different ways to approach the problem, depending on which "universe" of arrangements you are considering: people distinguished/undistinguished, same for seats, same for tables, plus whether the tables are round or linear.   
The wording of the problem might be interpreted as un-dinstinguished people, un-d. seats, distinguished tables of unspecified type. That is: how to put 18 balls in 4 distinguished boxes, with no less than 4 balls in each.
If that is the correct interpretation, then two balls are free for being allocated wherever: $4 \cdot 4 /2$ ways.
When this happens in a test at school or elsewhere, it is advisable to accompany the answer with the "interpretation".  
--- Addendum  --- 
There is some confusion going around here.
Let repeat @lulu's request and mine that when speaking of "ways" it is fundamental to specify which is the universe of equi-probable events being considered.
Let me try and clarify with an example.
In how many ways can you flip a pair of fair coins ?
a) we consider only the cases "equal outcome" (HH or TT) and "different"(HT), because the coins are not labelled and we cannot distinguish (HT) from (TH). They are equi-probable, on the basis physical considerations.
b) three cases (HH), (TT), (HT): the coins are again not distinguishable; however three results are not equi-probable.
c) the coins are labelled and we distinguish the four cases (HH),(HT),(TT) and (TH).
Then are the "ways" 2, 3 or 4 ?
Note that, when we come to compute the probability as No. of favourable results / Tot. number of results, and provided the results are weighted appropriately, for the probability of having "equals" we get $1/2$ in every case.
Now, for the problem posed by OP, there so many possible assumptions that can be made, that an answer in terms of "ways" cannot be given unless carefully specifying the "universe".
A: Using the fact that:

n people can be seated on a round table in (n-1)! ways.
Proof:
  All n seats on a table are alike. Select any person to be seated first. He has only 1 choice of seats (there is only 1 way, whichever seat he chooses).
   Now, symmetry is broken. So, remaining (n-1) seats can be considered as being in a line, with respect to the seat which was first occupied.
    So, arrange remaining (n-1) people in a line in (n-1)! ways.

There are 2 possible arrangements: $$4,4,4,6$$ and $$4,4,5,5$$
For, $4,4,4,6$: divide 18 people in these groups:
$$^{18}C_4\cdot^{14}C_4\cdot^{10}C_4\cdot^{6}C_6$$
seat them in: $$3!\cdot3!\cdot3!\cdot5!$$
But, one table  has 6 people while, other 3 have 4 each, so, select that one table in $^4C_1$ ways.
For, $4,4,5,5$: divide 18 people in these groups:
$$^{18}C_4\cdot^{14}C_4\cdot^{10}C_5\cdot^{5}C_5$$
seat them in: $$3!\cdot3!\cdot4!\cdot4!$$
2 tables have 4 while other 2 tables have 5 people. Select 2 out of 4 tables in $^4C_2$ ways.
$$(^{18}C_4\cdot^{14}C_4\cdot^{10}C_4\cdot^{6}C_6)(3!\cdot3!\cdot3!\cdot5!)(^4C_1) +(^{18}C_4\cdot^{14}C_4\cdot^{10}C_5\cdot^{5}C_5)(3!\cdot3!\cdot4!\cdot4!)(^4C_2)$$ is the final solution.
A: Given there are $ 4  $ different tables and $18$ people to be seated with at least $4$ in each  table.
Let us number the tables $ 1,2,3,4$ as all tables are different.
if we put at least $ 4 $ in each table we have $ 2 $ extra persons.
Here we have only 2 possible configurations:-


*

*$(6,4,4,4)$

*$(4,4,5,5)$
For the first one:-
$${^{18}C_6\times^{12}C_4\times^{8}C_4\times^{4}C_4\times{3!}^3\times{5!}\times{4}}$$
Here there are 4 different configurations of the selecting the table, ${\frac{4!}{3!}}$=$4$.
For the second case , $4,4,5,5$:-
$${^{18}C_4\times^{14}C_4\times^{10}C_5\times^{5}C_4\times{3!}^3\times{4!}\times{6}}$$
Here there are 6 different configurations of the selecting the table, ${\frac{4!}{2!\times2!}}$=$6$.
The result is the sum of the above two cases.
