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# Mathematics

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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

#1
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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
+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. tuffla2022 Aug 3, 2022