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Thread: 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

    $\displaystyle n! = n(n - 1)!$

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


    By letting $\displaystyle n = 1$ we have

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

    $\displaystyle 0! = 1$.
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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)?
    $\displaystyle n!$ is the number of permutations of the set $\displaystyle \{1,\ldots,n\}$, i.e. the number of maps from $\displaystyle \{1,\ldots,n\}$ to itself that are bijective.

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

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

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

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

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

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