15 different books to choose from a door prize winner can choose 4

Question

how many ways can the student select the 4 books and then arrange them on a shelf?

Answer 1

Hi Layne,

 

Whenever we deal with a problem like this, I think it helps to first decide whether we're talking about "combinations" or "permutations."

 

A "combination" problem is one where the order that the choices are arranged does not matter - such as selecting people for a committee. If I pick persons "A B C D", that group is the same as if I had picked "D C B A" - if I am only concerned with the overall group of people, then the order that they are selected does not matter.

 

A "permutation" problem is one where the order DOES matter, such as picking the 1st - 4th place winners of a contest. In this case, picking "A B C D" is different from picking "D C B A" - in this case, the order matters.

 

So, for the problem you listed, we want to come up with the number of ways that 4 books (out of 15) can be ordered on a shelf. Since the order of the books DOES matter, this is a permutation problem. Permutations are actually a little bit easier to solve than combinations - all we have to do is figure out:

 

   (# of possibilities for the 1st book) x (# of possibilities for the 2nd book) x (# of possibilities for the 3rd book) x (# of possibilities for the 4th book)

 

Each time we "select" a book, it reduces the number of possibilities for the remaining books - so:

 

   (15) x (14) x (13) x (12) = 32760.

 

There are 32760 different permutations for how a selection of 4 books (out of a possible 15) can be arranged on a shelf.