The number of ways to select a items out of b items is a! / b! (a -b)!, right?
So, if a = b, the answer has to be 1. (There's only one way to select everything.) Therefore a! / a! (a - a)! = 1, or 1/(0!) = 1. Hence, 0! = 1.
* * * * *
Or, how about this one...?
n! = n * (n - 1)!, by definition. Letting n = 1, then 1! = 1 * 0!. Since 1! = 1, this means 1 = 1 * 0!, and again 0! = 1
2007-01-09 15:13:15
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answer #1
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answered by Anonymous
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It's defined as 1. Some people long long ago decided that's what it should be. Honestly! While there are now proofs, those are all after-the-fact.
Why was it defined that way? The most likely theory is that a factorial represents the number of sets that can be made with the number of items.
1! = 1 set => { 1 }
2! = 2 sets => { 1, 2 } { 2, 1 }
3! = 6 sets => { 1, 2, 3 } { 1, 3, 2 } { 2, 1, 3 } { 2, 3, 1} { 3, 1, 2 } { 3, 2, 1}
0! has one set. What set you may ask? The empty (or 'null') set.
There's no ASCII character for it, but it looks like a big O with a slash going through it.
2007-01-09 23:17:39
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answer #2
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answered by slights 2
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Because it turns out that the Gamma Function, which is a continuous function, happens to be equal to n! when x = any integer n+1. The symbol for the Gamma function looks like F without the middle bar, so the following is true:
F(n+1) = n! where n is an integer
F(0+1) = 0!
but F(1) = 1
Therefore 0! = 1
So, that's why.
Addendum: Atheist has added a bit more useful facts to this story.
2007-01-09 23:13:53
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answer #3
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answered by Scythian1950 7
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the factorial is a special case (0 and positive integers) of the gamma function.
gamma(1) = 0! = 1
{In mathematics, the Gamma function is an extension of the factorial function to complex and non-integer numbers. The factorial function of an integer n is written n! and is equal to the product n! = 1 Ã 2 Ã 3 Ã ... Ã n. The Gamma function "fills in" the factorial function for non-integer and complex values of n. If z is a nonnegative integer, then
gamma(z + 1) = z! }
2007-01-09 23:08:14
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answer #4
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answered by atheistforthebirthofjesus 6
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Why does 0! = 1? It's just DEFINED that way.
Here are two (apparently) separate arguments why 0! is defined to be 1, though if you really know each of these situations you will realize that the mathematics is in fact really the same thing, and that it's actually exactly the same argument applied in two different contexts.
1. Since the earliest days that the concept of "n factorial" was arrived at,*** there have been various formulae involving those factorials, used in problems such as how to choose sets of 'r' objects from a total of 'n' objects, WITHOUT REGARD TO THE ORDER IN WHICH THEY'RE CHOSEN. (That is saying that ordered, successively chosen groups like [a, b and g], or [g, b and a] or [b, g, and a], etc. --- there are in fact 6 different ordered ways in which you could have picked the members of this one SET --- ARE in fact just ONE set with members a, b and g picked in any order, for these purposes.)
It was quickly realized that the formula for the number of ways that this could be done is written, using factorials, as:
n! / [r! (n-r)!]
We know what this means when r is not equal to n. However, in order to have it make sense when r = n, we simply have to define 0! as 1.
2. You can also see that this is desirable when you consider the general coefficient of x^r in in something like (1 + x)^n. It is also given by exactly the same formula shown above. You know that if n is an integer, this expansion closes, i.e. ends after a finite number of terms. It clearly starts with 1 and ends with (1) x^n; so both the first and last coefficients are 1. In order to get this from the general formula, one must define 0! to be 1 once again.
I suspect that this was realized as being highly desirable long before the Gamma function was invented. (How and why? : Because one of the most important and early applications of "choosing r objects from n without regard to order" was in card games played in royal courts. THOSE COURTIERS etc. were MUCH MORE INTERESTED in the practical outcomes when betting on their games, and the probabilities of various kinds of hand arising, than in the possibility that some mathematician in the future might come along and generalize the whole concept of factorials to non-integer values of n and r. I recall being told in my schooldays the names of one or two well-known mathematicians who were employed in, and highly regarded at courts for, calculating the probabilities of very many numbers of possible dealing outcomes, with or without knowledge of what had been drawn before. (It was extremely important for the betters to know such odds.) That's when I understand that formulae like the above --- and much more complex ones -- were first developed extensively; and in many of them, the need to continue to still make sense, when some factor would look formally like 0!, would require that the latter should be defined to be 1.)
Fortunately what worked for the integers and 0 in r! agreed nicely with things like the general recursive relations for both n! and for its later generalization, the Gamma function, so everyone continued to remain happy with the notion that 0! = 1.
Live long and prosper.
2007-01-09 23:12:26
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answer #5
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answered by Dr Spock 6
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The same questions keep popping up over and over! I see others have given you good informative answers to this, so I can't resist repeating a joke I heard recently here in this forum...
why is 0! = 1?
because if 0 turns into 1, it would be very surprised.
2007-01-09 23:19:57
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answer #6
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answered by Joni DaNerd 6
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It's a convenience to allow for simplicity of use.
Since n! = n*(n-1)! and 1! = 1, then you can define 0! =1 and the recursion relation remains valid for all positive n.
2007-01-09 23:15:50
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answer #7
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answered by modulo_function 7
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"n! is the number of permutations (ways of arranging) exactly n things. It makes sense to say that there is 1 way to arrange zero things."
2007-01-09 23:16:33
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answer #8
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answered by Sheen 4
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