Question

Factorial: 0!

Asked by: DrWarezz

Hello again :-)

Why does: 0! = 1 ??

Surely:

4! = 4*3*2*1 = 24
3! = 3*2*1     = 6
2! = 2*1         = 2
1! = 1             = 1
0! = 0

???????????????????

Thank you,
[r.D]

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Asked On
2005-01-28 at 08:32:34ID21292728
Tags

factorial

,

0

Topic

Math & Science

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Answers

 

by: SteHPosted on 2005-01-28 at 08:44:51ID: 13164950

Convention!

 

by: DrWarezzPosted on 2005-01-28 at 08:48:54ID: 13165002

..elaboration please? :-)
I've heard of Convention (can't remember anything about it though lol), but how does that prove that !0=1 ?

Thanks
[r.D]

 

by: ozoPosted on 2005-01-28 at 08:49:51ID: 13165008

n! = n*(n-1)!
1! = 1*0!

 

by: ozoPosted on 2005-01-28 at 08:55:59ID: 13165086

Because the number of permutations of 0 elements is 1

 

by: DrWarezzPosted on 2005-01-28 at 08:56:22ID: 13165090

Ah k. I understand what you've said.. but, then doesn't that contradict with:

0! = 0*(0-1)!

But, 0 * [anything] is always zero.. I just don't see the logic when it's put like that (0! = 0*(0-1)!).

ta,
[r.D]

 

by: crnimajloPosted on 2005-01-28 at 08:58:28ID: 13165120

http://mathworld.wolfram.com/Factorial.html

Factorial is defined for positive integers. You could never have (-1)!.

 

by: JR2003Posted on 2005-01-28 at 08:59:51ID: 13165145

It means that formula that use factorial workout nicely like the ones for Permutations and combinations, when n = k they still  arrive at the correct answer.

                n!
n_P_k = --------  
             (n - k)!


                 n!
n_C_k = ----------
             k!(n - k)!


Just out of interest:
An extension of the Factorial function is the Gamma function.

http://mathworld.wolfram.com/GammaFunction.html

Gamma(n) = (n - 1)!

It might be for this reason too that Factorial 0 is defined as 1.

 

by: ozoPosted on 2005-01-28 at 09:00:36ID: 13165155

Because G(z) = (n-1)!
          inf.
   G(z) = Integral x^(z-1) e^(-x) dx
           0

and G(1) = 1

 

by: DrWarezzPosted on 2005-01-28 at 09:06:11ID: 13165242

Surely 0! is an invalid operation (well, obviously not, but why?). Because:

"The factorial n! is defined for a positive integer n"

However:

"The positive integers are the numbers 1, 2, 3,"

..?

[r.D]

 

by: DrWarezzPosted on 2005-01-28 at 09:19:16ID: 13165431

>"The factorial n! is defined for a positive integer n"

However, my calculator works with negative numbers without error..? I'm confused.
I can only understand contradictions here.. :o\

Please help. :-)

[r.D]

 

by: ozoPosted on 2005-01-28 at 09:24:21ID: 13165485

0! is 1
(-1)! is undefined

 

by: DrWarezzPosted on 2005-01-28 at 09:24:35ID: 13165493

Oh!!! I understand it now!!

It was Ozo's comment: "Because the number of permutations of 0 elements is 1" that helped me figure it out. ^_^

Thanks all!
[r.D]

 

by: ozoPosted on 2005-01-28 at 09:27:22ID: 13165528

 

by: DrWarezzPosted on 2005-01-28 at 09:33:18ID: 13165601

Thank you all -- it was the word "permutation" that explained it all to me. ^_^ (Just so that the next time another idiot comes along and says "how to use factorial", or "why 0!=1", just says the word "Permutation") ^_~

Thanks,
[r.D]

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