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Math Help - 0!=1

  1. #1
    Newbie AfricanBorn's Avatar
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    Exclamation 0!=1

    I've got a very simple question yet it has been bugging me for ages, WHY/HOW does 0 permutations/arrangements =1 (0!=1)?
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  2. #2
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    It's simply a matter of definition - and it is a useful definition at that.


    But if you wanted to be really clever, note that

    n! = n(n - 1)!

    (n - 1)! = \frac{n!}{n}.


    By letting n = 1 we have

    (1 - 1)! = \frac{1!}{1}

    0! = 1.
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  3. #3
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    Quote Originally Posted by AfricanBorn View Post
    I've got a very simple question yet it has been bugging me for ages, WHY/HOW does 0 permutations/arrangements =1 (0!=1)?
    n! is the number of permutations of the set \{1,\ldots,n\}, i.e. the number of maps from \{1,\ldots,n\} to itself that are bijective.

    If n=0, then \{1,\ldots,n\}=\emptyset, and there exists one map from \emptyset to \emptyset... the empty map \emptyset!!

    Remember how maps are defined: a map from A to B is a subset f of A\times B such that, for every x\in A, there exists exactly one y\in B such that (x,y)\in f, and we write f(x)=y.

    If A=\emptyset, then f=\emptyset is a subset of A\times B(=\emptyset) (whatever B is) that satisfies the assumption since there is no x\in A (hence the condition is automatically fulfilled: it is empty). Thus, \emptyset is a map from \emptyset to anything. And it is the only one.

    It is bijective from \emptyset to \emptyset: for every y\in\emptyset, there is a unique x\in\emptyset mapped to y. Indeed, there is no such y, so there's nothing to be checked.

    Thus, \emptyset is the only map from \emptyset to itself, and it is bijective. As a conclusion, 0!=1.

    This makes sense in many ways, like Prove It illustrated.
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