20 books 5 different shelves So I'm trying to answer this question and am not sure if my answer is correct. In the text book I'm using, this question asked before combinations are even introduced (only permutations) so I'm not entirely sure if I'm in the right direction. Is it possible to answer these using permutations?

Consider a bookcase with 6 shelves and suppose that we have 15 different books to place on the shelves. How many different ways are there to place the books on the shelves if the left to right order on each shelf is unimportant?

The order being unimportant implies it's a combination question, and having different "containers" lead me to use the stars and bars approach.

$C(n + k -1, k) = C(5 + 15, 15) = C(20,15)$ ways to put the puts on the shelf when order is unimportant. 

The next part of the question says the order is important, assuming the above answer is right, would the number of ways with order simply be $15! * C(20, 15$)? Furthermore if each shelf is to get at least one book, is C(14, 9) correct when order is unimportant?

Some confirmation or correction would be appreciated, as would alternate methods for solving. Thanks!
If the left-to-right order on each shelf is unimportant, there are $6^{15}$ ways to place the books (as each book should be assigned with its shelf number).

If the left-to-right order is important, you have first order all books in the left-to-right and top-to-bottom order, and then place 5 "containers separators" between them (consider the similar task of writing up 15 different printable characters in a text file containing no more than 6 lines - i.e. no more than 5 newline characters). The answer would be $15! \times \binom{20}{5}$.
combination 

$nCr = \dfrac{n!}{(n-r)r!}$

Here $n=$ factorial of no of units/items 

$r=$ chance no

example $20$ similar books be placed on $5$ different selves...

$\dfrac{20!}{(20-5)5!}$