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Re: 0! = 1
Posted:
Aug 9, 2001 11:58 PM


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 wellknown:
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 ÃÂÃÂ· (n1)! > > > > 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 > > >



