Originally Posted by

**Pinkk** Not really probability/statistics but this is an introductory discrete mathematics class and I'm rusty on all of this material, so here it goes:

Pamela has 15 different books. In how many ways can she place her books on two shelves so that there is at least one book on each shelf?

So I considered the one event where 14 books are on Shelf A and only 1 on Shelf B, and so the possible arrangement of books in that scenario is $\displaystyle 14! \times 1!$. Now, if there are 13 books on Shelf A and 2 are on Shelf B, the number of possible arrangements are $\displaystyle 13! \times 2!$. And so it goes on as such and so the total possible arrangements would be $\displaystyle 2(14! \times 1! + 13!\times 2! + ... + 9! \times 6! + 8! \times 7!)$. Is this correct or am I just way off? Thanks.