1. Combinatory

I have a lot of doubts on how to solve this problem.
How many different ways can I put 12 books into 4 different bookshelves if all the books are different and the position of the books in each bookshelf is important.

2. Re: Combinatory

Originally Posted by Yeison
I have a lot of doubts on how to solve this problem.
How many different ways can I put 12 books into 4 different bookshelves if all the books are different and the position of the books in each bookshelf is important.
I always end up getting these frigging combinatorics problems wrong but what I think the answer here is is

$12! 4^{12}$

You first select a book from the 12 and put it on one of 4 shelves. There are 12x4 ways to do this.
Then you select a book from the 11 and put it on one of 4 shelves. There are 11x4 ways to do this.
And so on. The final product is

$(12\cdot 4)(11 \cdot 4) \dots (1 \cdot 4) = 12! 4^{12}$

3. Re: Combinatory

Originally Posted by romsek
$12! 4^{12}$
That is correct if we assume that each book shelf can hold all twelve books.

4. Re: Combinatory

Hello, romsek!

You first select a book from the 12 and put it on one of 4 shelves.
. . There are 12x4 ways to do this.

Then you select a book from the 11 and put it on one of 4 shelves.

And place it to the right of the previous book.
. . There are 11x4 ways to do this.

And so on.

The final product is: .$(12\cdot 4)(11 \cdot 4) \dots (1 \cdot 4) = 12! \:\!4^{12}$

Brilliant!

Since you arrived last November,
. . you have provided many elegant solutions.

A belated "Welcome board", Romsek!