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0! = 1 (1 Viewer)

Iruka

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Its true by definition.

They define it that way so that most of the formulae using factorials will still work out nice.

Alternately, you can think of n! being the number of ways to arrange n distinct objects in some order. In which case, if n=0, there is only one way to arrange no objects in order - i.e., not at all.
 

DJel

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Stevo. said:
Good work... :rolleyes:

Iruka said:
Its true by definition.

They define it that way so that most of the formulae using factorials will still work out nice.

Alternately, you can think of n! being the number of ways to arrange n distinct objects in some order. In which case, if n=0, there is only one way to arrange no objects in order - i.e., not at all.
Thanks for that. I was unable to find a definitive proof and assumed this was so the formulas would work out.
 

Poad

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The actual proof is:

n! = n(n-1)!
let n = 1

1! = 1(1-1)!
.:. 1 = 0!
 

Templar

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Poad said:
1! = 1(1-1)!
.:. 1 = 0!
If we're going to be completely rigorous here, the last line doesn't follow from the above, since then you are assuming 1!=1 and that n! is defined to be n(n-1)! even for n=1.
 

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