If I have 12 books and 12 book spaces, how many ways can I arrange these books? Not all spaces have to be filled. All the books are the same. If I have 12  books and 12 book spaces, how many ways can I arrange these books? Not all spaces have to be filled. All the books are the same. In other words, putting a book in space 1 and a book in space 3 and leaving spaces 2, 4, 5, 6, 7, 8, 9, 10, 11, 12 empty is a valid combination. 
 A: There are just as many ways as there are subsets of the $12$-element set of book spacess. And a $12$-element set has $2^{12}$ subsets. 
To see this another way, imagine the spaces (slots) are arranged in a row. We start at the left end and have $2$ choices, put in a book or not. For every choice about the first slot, there are $2$ choices for the second slot, book or no book. And so on. So there is a total of $2\times 2\times\cdots\times 2$ (length $12$ product) choices, that is, $2^{12}$ choices. This includes the choice of saying no $12$ times, that is, placing no books.
A: Since all the books are the same, you only have to choose which spaces are filled (and all the unused books are left out).  There are $2^{12}$ ways to choose a (possibly empty) subset of $12$ spaces.
A: Another way is, to calculate the number of combinations, if k spaces are left. It is $12 \choose k$
Now you have to sum the combinations for k from 0 to 12.
$\sum_{k=0}^{12} {12 \choose k}$ 
According to the Binomial Theorem it is $\sum_{k=0}^{12} {12 \choose k}=2^{12}$
If at least one book have to be placed in the bookshelf, then the number of combinations are $\sum_{k=0}^{11} {12 \choose k}=\left[ \sum_{k=0}^{12} {12 \choose k}\right] -{12 \choose 12}=2^{12}-1$
