#2**+10 **

0! is defined to be 1 so that certain formula work.

For example: the formula for permutations is: _{n}P_{r} is n! / (n - r)!

which wouldn't work for _{5}P_{5} = 5! / (5 - 5)! = 5! / 0! unless 0! is defined to be 1.

A problem that requires _{5}P_{5} 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!.

geno3141
Mar 2, 2015

#1**+10 **

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

CPhill
Mar 2, 2015

#2**+10 **

Best Answer

0! is defined to be 1 so that certain formula work.

For example: the formula for permutations is: _{n}P_{r} is n! / (n - r)!

which wouldn't work for _{5}P_{5} = 5! / (5 - 5)! = 5! / 0! unless 0! is defined to be 1.

A problem that requires _{5}P_{5} 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!.

geno3141
Mar 2, 2015

#3**+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.

Melody
Mar 3, 2015

#4**+5 **

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"

CPhill
Mar 3, 2015