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# Why is 0!=1?

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Why is 0!=1?

\$\${\mathtt{0}}{!} = {\mathtt{1}}\$\$

Guest Mar 2, 2015

#2
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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!.

geno3141  Mar 2, 2015
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#1
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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
+17711
+10

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!.

geno3141  Mar 2, 2015
#3
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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
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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
#5
+91797
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