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Jason has 4 books lined up for reading over the holiday break. How many choices does he have for determining the order in which he can read the books?

 Aug 3, 2022

Best Answer 

 #1
avatar+113 
+2

Is we assume these books are all different, we can call the books A, B, C and D.

You can start lining up the possibilities. Let's start with book A:

A -> B -> C -> D

A -> B -> D -> C

A -> C -> B -> D

A -> C -> D -> B

A -> D -> B -> C

A -> D -> C -> B

There are 6 different ways to go through these books starting with book A.

The same is true if we start with any other book. So there are 6 x 4 = 24 choices for the order he can read the books.

 

We use a formula for this kind of permutations problem:

Number of permutations: n! (n faculty) = n * (n-1) x (n-2) x ... x 2 x 1

 

For only 4 books there are 4 x 3 x 2 x 1 = 24 choices.

 

For 6 books there would have been: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 choices.

 

 Aug 3, 2022
 #1
avatar+113 
+2
Best Answer

Is we assume these books are all different, we can call the books A, B, C and D.

You can start lining up the possibilities. Let's start with book A:

A -> B -> C -> D

A -> B -> D -> C

A -> C -> B -> D

A -> C -> D -> B

A -> D -> B -> C

A -> D -> C -> B

There are 6 different ways to go through these books starting with book A.

The same is true if we start with any other book. So there are 6 x 4 = 24 choices for the order he can read the books.

 

We use a formula for this kind of permutations problem:

Number of permutations: n! (n faculty) = n * (n-1) x (n-2) x ... x 2 x 1

 

For only 4 books there are 4 x 3 x 2 x 1 = 24 choices.

 

For 6 books there would have been: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 choices.

 

tuffla2022 Aug 3, 2022

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