0! is defined to be 1 so that certain formula work.
For example: the formula for permutations is: nPr is n! / (n - r)!
which wouldn't work for 5P5 = 5! / (5 - 5)! = 5! / 0! unless 0! is defined to be 1.
A problem that requires 5P5 is: How many different ways can you arrange 5 books on a bookshelf?
Since there are 5 ways to choose the first book, 4 ways to choose the second book, 3 ways to choose the third book, 2 ways to fourth books and 1 way to choose the last book, you get 5 x 4 x 3 x 2 x 1 = 120 = 5!.
Not a proof....but notice....
3! = 4!/4
2! = 3!/3
1! = 2!/2
So....this would imply that
0! = 1! / 1 = 1/1 = 1
0! is defined to be 1 so that certain formula work.
For example: the formula for permutations is: nPr is n! / (n - r)!
which wouldn't work for 5P5 = 5! / (5 - 5)! = 5! / 0! unless 0! is defined to be 1.
A problem that requires 5P5 is: How many different ways can you arrange 5 books on a bookshelf?
Since there are 5 ways to choose the first book, 4 ways to choose the second book, 3 ways to choose the third book, 2 ways to fourth books and 1 way to choose the last book, you get 5 x 4 x 3 x 2 x 1 = 120 = 5!.
Thank you all.
This is a good question from an enquiring mind.
I liked both answers, the 2 answers compliment each other really well I think.
As an addendum to geno's explanation.....suppose we don't want to choose any books from the shelf.....in other words...C(5,0)
The "formula" is
5! / [(5 - 0)! * 0!] = [5!/ (5! * 0!)] = [1 / 0! ]
Well.....there 's only one way not to choose any books from the shelf.....don't take any of them!!!
Thus......the 0! in the "formula" MUST = 1 for this to "work"