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All of us know that 0! is 1 but i just am eager enough to know "how"? Also, does 0! really "EXIST"? Means, if it does then why not the factorials of -ve numbers?
How can one "DEFINE" it??
Last edited by ZHero (2008-06-21 08:39:56)
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When thinking about combinations and permutations, it makes sense that 0! = 1.
The number of way of choosing r objects from a choice of n is nCr, or
If we use 0!=1, then the number of ways of picking no objects turns out to be
Another way of rationalising it is by considering this pattern:
3! = 4!/4
2! = 3!/3
1! = 2!/2
∴ 0! = 1!/1 = 1/1 = 1.
If you try to continue that to the negatives, then you'd get that (-1)! = 0!/0, which isn't allowed.
All factorials of other negative integers involve (-1)! in their definition (possibly very indirectly), and so no negative integer has a factorial.
You might be interested in the Gamma function, which is an "expansion" of factorials.
This relates to factorials because:
The Gamma function of 1 is equal to 1, giving another argument to why 0! = 1.
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Oh! That was such a beautiful explanation 'mathsyperson'! I bet i could not have known this from anywhere else!
This definitely makes Great Sense!
Many a Thanks!!
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mathsy: you should write pages for mathsisfun.com, that was pretty good.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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also note that by using the gamma function you can define the factorial for any non-negative-real-integer over the whole complex domain.
http://mathworld.wolfram.com/images/eps-gif/Factorial_1000.gif
http://mathworld.wolfram.com/images/interactive/FactorialReImAbs.gif
Last edited by luca-deltodesco (2008-06-21 19:23:32)
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