Permutations with forbidden positions - sit $n$ students in a row 
Suppose we have $N$ students $(N>20)$ that should be seated in a row. ($1-n$
  seats)
$1.$ In how many ways can we arrange such that student number $2$ won't sit on chair number $1,3$ or $7$?
$2.$  In how many ways can we arrange such that student number $2$ won't sit on chairs number $1,3$ or $7.$ Student number $1$ won't sit on chairs
  $1,2$ or $3$ and student number $7$ won't sit on chair number $13$.

For question number $1$ my answer is $(n-3)\cdot(n-1)!$
But I'm getting troubles with question $24.
I can't see how to count it, should I use inclusion-inclusive-principle?
Any hints?
Thanks a lot. 
 A: Since you have done the first one I will go through number $2$. For the sake of brevity I will give the highlights of the method and provide links for further explanation.
I would use rook polynomials.
In this case the chess board looks like
$$\begin{array}{cc}
& \text{chairs} \\
\text{students} &
\begin{array}{c|c|c|c|c|c|c|c}
  &2&1&3&7&13&4&\cdots \\\hline
1 &\bbox[silver,10px]{\phantom{H}} &\bbox[silver,10px]{\phantom{H}} &\bbox[silver,10px]{\phantom{H}} & & & &\cdots \\\hline
2 & &\bbox[silver,10px]{\phantom{H}} &\bbox[silver,10px]{\phantom{H}} &\bbox[silver,10px]{\phantom{H}} & & &\cdots \\\hline
7 & & & & &\bbox[silver,10px]{\phantom{H}} &\bbox[white,10px]{\phantom{H}} &\cdots \\\hline
3 &\bbox[white,10px]{\phantom{H}} & & & & & &\cdots \\\hline
4 &\bbox[white,10px]{\phantom{H}} & & & & & &\cdots \\\hline
\vdots & \vdots &\vdots &\vdots &\vdots &\vdots &\vdots&\ddots \\
\end{array}
\end{array}$$
We have two disjunct forbidden subboards, one consisting of a single square, this has rook polynomial
$$1+x$$
the other consists of two rows of $3$ squares with the lower $3$ offset by $1$ square. The rook polynomial for this is found by manually counting the placements of identical non-attacking rooks. There is $1$ way to place $0$ rooks, $6$ ways to place $1$ rook and $7$ ways to place $2$ rooks, hence the polynomial for this subboards is
$$1+6x+7x^2$$
since they are disjunct subboards the total rook polynomial is
$$(1+x)(1+6x+7x^2)= 1 + 7x + 13x^2 + 7x^3$$
Since the board is $N\times N$ with $N\gt 20$ (assuming the same number of students as seats) then this polynomial transforms by replacing 
$$x^k\longrightarrow (-1)^k(N-k)!$$
giving our final answer:

$$\text{arrangements}=N!-7(N-1)!+13(N-2)!-7(N-3)!\tag{Answer}$$

Please see my answer here for more on rook polynomials.
A: You are correct for your answer to the first part of the question.
For the second part of the question, I would indeed approach with inclusion-exclusion.
Let $A$ be the set of arrangements for student2 does sit on chair $1,3,$ or $7$.  $B$ the set of arrangements where student1 does sit on chair $1,2,$ or $3$, and $C$ where student7 does set on chair $13$.
We want to count $|A^c\cap B^c\cap C^c|=|\Omega|-|A\cup B\cup C| $
$= |\Omega|-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$
For calculating each of these terms, approaching directly seems easiest.  When calculating one of the terms where $A$ and $B$ simultaneously happen, suggest breaking into cases based on whether student $1$ is sitting in one of the desired chairs for student2 or not.

 Calculating $|A\cap B\cap C|$, pick student7's seat to be chair13, then pick which of 1,2,3 student1 sits in.  If seat2 was chosen, then pick which of the three chairs 1,3,7 student2 sits in, else if seat1 or seat3 were chosen, pick which of the two remaining chairs from 1,3,7 student2 sits in.

