Answer:
[tex]9.332622\times 10^{155}[/tex]
Step-by-step explanation:
Before getting the final result, let's analyse the same situation but with less books for having a better perspective. For example, just 3 books, named: A, B and C
3 books mean 3 spaces
At the beginning, the first space has 3 ways to be occupied: they are book A, B or C.
If the first place is already occupied for any of them, then, the second place could be occupied just by the two others. If the second place is occupied by any of those two, finally, the third space just could be occupied by one.
Imagine that first space is occupied by book A. The ways to arrange the other books would be:
As you see, there are two ways to arrange them positioning the book A at the beginning. The same would happen with book B and book C at the beginning. It means there are 3 options at the beginning and then just 2.
If we multiply those values we get the total ways. It means 6 ways.
Real Problem
Now, for the 99 different books the analysis would be the same:
At the beginning would be 99 books to put in the first space, then in the second space would be 98, then 97, then 96, and so on, until 1.
But it is not a problem because there is a way to write it. It is called factorial.
[tex]99! = 99\times 98\times 97\times 96\times ...\times 3\times 2\times 1[/tex]
If we make all this multiplication we get [tex]9.332622\times 10^{155}[/tex] ways to order the 99 books.
