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Why is 0! = 1 ??? (1 Viewer)

DA

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The factorial is defined in terms of Euler's gamma function, which is an integral. The integral is

Gamma(n) = int x^(n-1) e^-x over x = 0 to x = inf.

and it can be shown that Gamma (n) = (n - 1)!.

So put n = 1 and the integral is the integral of e^-x over x = 0 to x = inf. is 1, which gives us

Gamma (1) = 1

or 0! = 1.
 

McLake

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I was under the impression that was a definition (in other words, "we define ! to mean ... and 0! = 1")
 

wogboy

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5! = 120
4! = 5!/5 = 24
3! = 4!/4 = 6
2! = 3!/3 = 2
1! = 2!/2 = 1
0! = 1!/1 = 1

See the pattern? The principle used here is that n!/n = (n-1)! So if we let n = 1,

LHS = 1, RHS = 0!

So, 0! = 1
 

Weisy

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ah...I like that much better than the other one ;)
 

ravingcow

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The other way to look at it is this:

How many ways can you arrange 5 items? 5! = 120

How many ways can you arrange 1 item? 1! = 1

How many ways can you arrange zero items? Only one way, of course! :)

There being zero ways to arrange zero items doesn't quite make sense to me, although I am sure others out there will disagree.

Oh well.
 

ravingcow

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Okay, okay. So Dumbarse just mentioned that.

It was still a good point, though...
 

macca202

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9C3 = 9C6 (6+3=9) etc.., ah no this is going nowhere...
9C9 which = 9!/9!x(9!-0!), oh shit i just totaly confused myself... =9!/9!x(9-9)! ah here we go...
=9!/9!x0!
obviously only 1 combination possible, so 0!=1
 

wogboy

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Factorials are just a miserable attempt at making maths look exciting :p
 

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