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?
+113 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.
+113 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.
