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!

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    $\begingroup$ Combinations are just a way of doing permutations and taking account of the repetitions due to ordering ($C(n,k)=P(n,k)/P(k,k)$). So anything that is being done with combinations can ultimately be justified using permutations alone. $\endgroup$ – Arturo Magidin Apr 24 '12 at 5:23

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 $15! \times \binom{20}{5}$.

  • $\begingroup$ Thanks for the reply, I was wondering if you could clarify the first part where you say there are 15^6 ways to place the books. I guess I'm having a mind blank, because I can't really see how you got there. Once one book is put onto a shelf, the remaining books decrease so I thought this would lead to a factorial of some kind. Thanks $\endgroup$ – jason Apr 24 '12 at 5:42
  • $\begingroup$ @jason I had a mistype in my original answer (which seems not to affect your comment). You have to assign each book with its shelf number. Basically, the way to place the books could be written as a sequence of 15 digits, where each digit is from $1$ to $6$. You're taking the book number $1$ and are saying "I will place it to the $a_1$th shelf"; and then you're taking the book number $2$ and are saying "I will place it to the $a_2$th shelf"; and so on, where $a_i$ is from $1$ to $6$. There is obviously $6^{15}$ ways to do this. $\endgroup$ – penartur Apr 24 '12 at 5:46
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    $\begingroup$ Ahh, that clears it up. Thanks a lot :) Is there a way to mark your answer is the correct one? Sorry, I'm new here. Edit: Found it $\endgroup$ – jason Apr 24 '12 at 5:50


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

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

$r=$ chance no

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


  • $\begingroup$ Please see this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig May 4 '15 at 9:19
  • $\begingroup$ Your answer is incorrect. First of all, $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ Also, note that you first have to select which $k_1$ of the $20$ books are placed on the first shelf, which of the $20 - k_1$ books are placed on the second shelf, and so forth. $\endgroup$ – N. F. Taussig May 4 '15 at 9:20

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