Variations: in how many ways can people be arranged so I have this problem in my discrete-mathematics book that I have tinkered around for over a day at this point - in how many ways can 8 introverts sit on a row of 30 chairs so that there no 2 of them sit beside each other.
What would be the right way to aproach this kind of problem? 
It's in latvian but I have tried to arrange different ways how you cold seat them and so on, but I dont know how to include all the variations
All I can get to is that I need to calculate in how many ways those people can sit (which is variations of 8 from 30) and subtract all invalid positions (this is the part I can not figure out) 
 A: Hint: Consider $30-8+1=23$ cells and arrange there arbitrary 8 pairs $\bullet\times$ (person+chair) and 15 $\times$ (chairs). In the end remove the last $\times$ (chair). If bullets (people) are distinguishable multiply the result by $8!$.  
A: I would like to talk about the general methodology for solving this kind of problems.
As a general approach, start by defining the number $G(n,k)$ to be the number of ways to pick $k$ chairs out of $n$ with no two beside each other.
Then try to find recurrence relations between the numbers $G(n,k)$.
To count $G(n,k)$, we separate two cases. If the left most chair is not picked, then we may simply remove that chair, so there are $G(n-1,k)$ ways in this case.
Otherwise the left most chair is picked, hence the second left most chair must not be picked, and we have $G(n-2,k-1)$ ways in this case.
Therefore we get $G(n,k)=G(n-1,k)+G(n-2,k-1)$ whenever $n\geq 2$ and $k\geq1$.
The remaining task is to solve this and get $G(30,8)$. This can be computer-aided, or using various combinatorial tricks, such as generating functions.
Let $g_k(T)$ be the generating function $\sum_{n\geq0}G(n,k)T^n$. It's clear that $g_0=\frac 1 {1-T}$ and $g_1=T+2T^2+\dotsc =\frac T {(1-T)^2}$.
From the recurrence relation, for $k\geq 2$ we get $g_k= Tg_k+T^2g_{k-1}$, i.e. $g_k = \frac{T^2}{1-T}g_{k-1}$. A simple induction on $k$ then shows that $g_k = \frac{T^{2k-1}}{(1-T)^{k +1}}$.
Therefore we have $G(n, k)$ is the coefficient of $T^{n - 2k + 1}$ in $\frac 1 {(1-T)^{k +1}}$, which is $\binom{n-k+1}{k}$.
