In how many ways I can put books on a shelf when only colour matters I quite dont understand what is being asked in the question:
You have a lot of books, in 4 colours: red, blue, green and white. You want to put 20 books on a shelf. In how many ways can you do that if:


*

*Only number of books of a specific colour matter

*Only the sequence of colours on the shelf matter

*Only the sequence of colours matters and two books with the same colour cannot stand next to each other


I dont follow even the beginning. For part 1 - do I have to find in how many ways I can pick 20 elements from a set of 4 elements [types?] where order doesnt matter and the elements are not unique? (Red book 1 does not differ from red book 2)?
 A: All of these things have already been given in the comments – I am merely restating this in answer format.
The number of ways for part 1 is given by a multiset computation:
$$\left(\!\!\binom{20}{4}\!\!\right)=\binom{20+4-1}{4}=8855$$
The number of ways for part 2 is simply $4^{20}=1099511627776$. Part 3 is the same, but after placing the first there are only three choices for the rest, hence $4\times3^{19}=4649045868$.
Combinatorics problems are actually quite difficult to state in natural language, and rarely stated well (especially on this site). The relevance of order or repeats may be left to the reader, red herrings (for this question, what the contents of the books are) may creep in or the words might otherwise be misinterpreted and lead to an unintended result. This is why such problems are best stated/solved in their raw mathematical formulation. For your question:

In how many ways can a list of length 20 be filled by four distinct objects…
  
  
*
  
*…if the order is irrelevant?
  
*… if the order is relevant?
  
*… if the order is relevant, but the same object may not occupy two adjacent positions?
  

A combinatorics question I thought was well-stated can be found here.
