How to approach questions like "How many ..." I am currently studying for a big exam. In almost all of the topics I understand the basic ideas and how to approach them.
But not for combinatoric-, variance- and permutation questions. It does not matter how many questions I try to solve; I still make errors in every other question.
Consider for instance this question: 

How many seat arrangements are there in a class of 30 students if there are 15 tables (2 students per table) and two particular students must not sit together?

It would never have come to my mind that I have to substract $15*28!$ from $30!$. Now, after I have the solution it is somewhat logical. But me coming up with that solution? No way... And why isn't it $30!-2*15*28!$ since students could swap seats and still sit together?
As you can see, I am quite desperate and every advice would be welcome (including online resources). The exam is in ~10 days and I sincerely hope that I will understand this topic until then.
 A: Questions involving the word "not" are often best approached by addressing the complementary event. That is,
Number of Good Events = Number of Total Events - Number of Bad Events.
For your example, if we call the two students A and B, then 
Number of seatings with A and B not together = Number of unrestriced seatings - Number of seatings with A and B together.
The number of unrestricted seatings is $30!$, since we may place the $30$ students in any order.
To determine the number of seatings with A and B together, first pick a table for them in $15$ ways. Next, set them down at that table in $2$ ways. Finally, seat the remaining $28$ students in any of $28!$ ways.
Finally, we have
Number of seatings with A and B not together = $30! - 2 \cdot 15 \cdot 28!$.
A: You're given a fundamental problem:
"How many seating arrangements are there of 30 students in 15 tables"
and a restriction
"with two particular students not sitting together"
The idea is to first solve the bigger problem. With 30 students you have $30!$ ways of arranging them. So this is our preliminary answer. It doesn't satisfy the restriction though, so we need to deal with that.
A key here is to notice that your condition has a "not" in it. Counting how many configurations have the students not sitting together is fairly difficult. On the other hand, counting how many do have them sitting together is quite easy. Just sit them together and see what else you can do. If you seat the pair together, then you have $15$ tables to sit them at, $2$ ways for them to sit at any given table, and $28!$ ways to arrange the other students, leading to $2\cdot 15\cdot 28!$ ways where they are sitting together.
In the original problem we wanted the configurations where they weren't sitting together, so we have some work to do. But this ends up being pretty easy from this point. To find out how many don't have them sitting together, find out how many do and subtract that off of the total number of configurations. But we've already done both of those things! This leads to $30!-30\cdot 28!$ possible solutions.
The main ideas are these:
-split the problem into the broad configurations and the restrictions. Consider each separately.
-for each restriction you're given, consider it's negation as well. Pick the one of these to deal with that looks easier.
Once you can separate the problem out like this you can consider each term separately, which is much nicer than trying to derive that final number on its own.
A: Another way of seeing this : you construct a possible arrangement using a series of choices, where every arrangement can be obtained once and only once.
First we need to work on the question to be sure what it means. Here, we will assume that people switching chairs on a table yields to a different arrangement, and that people switching tables but respecting the original pairing yields to a different arrangement.
We need to seat A : 30 choices
then B : 28 choices since A's table is blocked
then the others : 28! choices.
So (simple multiplication) we have 30*28*28! possibilities. (it's the same as the other answers since 30! - 2*15*28! = (30*29 - 30)28! = 30*28*28! )
