- #1

- 321

- 20

## Homework Statement

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. (consider the books in each arrangement to be stacked one next to the other, with the first book on each shelf at the left of the shelf)

## Homework Equations

## The Attempt at a Solution

apparently the answer is (15!)(14) but I just don't see how.

so this is how I see it..

case 1:

one book on shelf L, 14 books on shelf R

case 2:

two books on shelf L, 13 books on shelf R

case 3:

three books on shelf L, 12 books on shelf R

case 4:

four books on shelf L, 11 books on shelf R

case 5:

five books on shelf L, 10 books on shelf R

case 6:

six books on shelf L, 9 books on shelf R

case 7:

seven books on shelf L, 8 books on shelf R

case 8:

eight books on shelf L, 7 books on shelf R

case 9:

nine books on shelf L, 6 books on shelf R

case 10:

ten books on shelf L, 5 books on shelf R

case 11:

eleven books on shelf L, 4 books on shelf R

case 12:

twelve books on shelf L, 3 books on shelf R

case 13:

thirteen books on shelf L, 2 books on shelf R

case 14:

fourteen books on shelf L, 1 book on shelf RI'll just go backwards I guess

for case 14, if you choose 1 of the 15 books and put it on shelf R, shelf L can have 14! combinations of the remaining books. Now if you put that book that was on shelf R and exchange it with a different book on shelf L, shelf L can have 14! new combinations of books because we switched the two books out.

You can do this for all 15 books

so just for case 14 you get a combination of: (15)(14!)

then for case 13, you have 2 books on shelf R, 13 on shelf L. shelf L can be arranged 13! ways while shelf R can be arranged 2! ways. so (13!)(2!). If we only switch out one book, and keep another, it will be (13!)(2!)(14) because there are 14 possible books that we can switch out while keeping one of the books.

Then there are (13!)(2!)(14)(15) combinations because there are 15 books total that you're going to useI'm just going to stop here. My answer leads me to adding all the the combinations in all of my cases, and just adding case 14 and 13 is already an extreme overestimate of the answer that the book has given me.

I just don't understand where my logic is failing though. Can anyone let me know?