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Re: 0! = 1
Posted:
Aug 9, 2001 11:58 PM
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I think the most logical reason I've heard for 0! = 1 is the following:
Consider Pascal's triangle.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
The rth term of the nth row is the well-known:
n!
nCr = (nPr)/r! = -------------
r! * (n - r)!
We know that the 0th term of _any_ row must be 1. When we set r = 0, we
have:
n!
-------
0! * n!
So now we want a value for 0! such that the above formula will evaluate to
1.
The obvious choice for 0! is 1. If we don't have 0! = 1, the formula only
holds for values 0 < r < n, which isn't as useful (also for the binomial
theorem).
-- Entropix
"The Scarlet Manuka" <sacha@maths.uwa.edu.au> wrote in message
news://9kvk32$a2d$1@fang.dsto.defence.gov.au...
> "michael" <michael@farheap.com> wrote in message
> news://622b25fd.0108091853.17c6b78@posting.google.com...
> > Why is 0! defined as 1? I have heard that it was done this way so that
> > the following recursive definition of factorial could be used.
> >
> > 0! = 1
> > n! = n ÃÂ÷ (n-1)!
> >
> > But why not just define 1! = 1 That would work just as well. I presume
> > there must be a better reason.
>
> We do define 1! = 1, but if we want the equation to work for n = 1 too
> we need 1! = 1.0! which implies that we should have 0! = 1.
>
> However, a better reason is that it is useful to have an empty
> product evaluating to 1, and 0! is an empty product. Similarly
> we usually want an empty sum to evaluate to 0.
>
> --
> The Scarlet Manuka
>
>
>
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