# Factorial of 0 - a convenience?

If I am correct in stating that a factorial of a number ( of entities ) is the number of ways in which those entities can be arranged, then my question is as simple as asking - how do you conceive the idea of arranging nothing ?

Its easy to conceive of a null element in the context of arrays, for example - so you say that there is only one way to present a null element.

But, in layman terms - if there are three humans h1, h2, h3 that need to be arranged to sit on three chairs c1, c2, c3 - then how do you conceive of a) a null human, and b) to arrange those ( that ? ) null humans ( human ? ) on the three chairs ?

Please note that referral to humans is just for easy explanation - not trying to be pseudo-philosophical. Three balls to be arranged on three corners of a triangle works just fine. So basically, how do you conceive of an object that doesn't exist, and then conceive of arranging that object ?

So, in essence ... is 0! = 1, a convenience for mathematicians ? Not that its the only convenience, but just asking. Of course, there are many. If yes, then its a pity that I can't find it stated like so anywhere. If not, then can anybody suggest resources to read actual, good proofs ?

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I've always felt comfortable with $0!=1$, because $(n-1)!=\frac{n!}{n}$ – balpha Mar 8 '11 at 17:45
possible duplicate of Prove $0! = 1$ from first principles – Aryabhata Mar 8 '11 at 17:47
errmm ... those are some convenient methods to prove that ... but without sounding pretentious ... I don't think those are the answers I am looking for. – alcy Mar 8 '11 at 17:53
@alcy: if you think you are asking a different question, you should specify what is different about your question. – Qiaochu Yuan Mar 8 '11 at 18:01
This is yet another duplicate. This question was asked yesterday: math.stackexchange.com/questions/25333/why-does-0-1/25341#25341 And here is another one: math.stackexchange.com/questions/20969/… (edit: second is mentioned by Moron) – Eric Mar 8 '11 at 18:06
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I don't see any difficulty with the idea of "arranging a set of zero things"... There is no need to

conceive of an object that doesn't exist, and then conceive of arranging that object

mostly, because that does not make much sense at all.

If I give you 2 apples and I tell you to arrange them in a line above the table, there are 2 possible outcomes. If I give you zero apples, how many possible possible things can I find on the table after you are done? Exactly one: no apples on the table.

By the way, every definition in math is done out of convenience.

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"By the way, every definition in math is done out of convenience." In that case, I am happy with this answer. – alcy Mar 8 '11 at 18:11
... but, again, you can NOT give me ZERO apples to arrange, in the first place. ( accidentally hit enter on my previous comment ) – alcy Mar 8 '11 at 18:13
Sure you can. I just gave you 0 apples to arrange. – Mitch Mar 8 '11 at 18:43
Well ... yes, all I wanted to know was whether its a convenience. The abstraction is now implicit. – alcy Mar 8 '11 at 18:59
@alcy: see math.stackexchange.com/questions/20969/…. Dealing with 'boundary' cases is hard because they often don't evoke real world examples so easily. So sometimes convenience is enough. A combinatorial interpretation can often give an answer from first principles (that is, not purely out of convenience...similar arguments works for $0^0 = 1$ and ${0 \choose 0} = 1$. On the other hand Mariano is right: in a manner of speaking -all- math is a matter of convenience. – Mitch Mar 9 '11 at 2:20
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There is exactly one way to arrange nothing: the null arrangement. You've misused your chair analogy: when you arrange null humans, you do it on null chairs, and there is exactly one way to do this.

Perhaps the following alternate definition will make things clearer. Suppose I have $n$ cards labeled $1, 2, ... n$ in order, and I shuffle them. Then $\frac{1}{n!}$ is the probability that they will stay in the original order. If $n = 0$, this probability is $1$ since $0$ cards can only be arranged in one possible order.

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The definition can be motivated by the $\Gamma$-function, but let's ignore that.

We can think of $0!$ as an empty product if we wish, the notion of "nothing" times "nothing" is $1$. But how do we arrive at this?

Instead of $1\times 2\times 3\times 4$, let's write $\times(1,2,3)$. I suppose you agree that we could break this up into arbitrary sub-products: $\times(1,2,3) = \times(\times(1,2),\times(3))$. Naturally, we can interpret $\times(3)$ to just be $3$. But suppose we broke this up even more: $\times(\times(3),\times())$. What would the most natural choice of $\times()$ be? Of course, $1$.

In the case of factorial, we can interpret $1!$ to be $\times(1, \times()) = 1\cdot(0!)=1$.

For other binary operations, we can determine the "empty"-operation by a simple rule-of-thumb: if we have the commutative operation $\oplus$ (just a generic operation), we can define the "empty-$\oplus$" as the value of $\iota$ such that $\forall x.\, \iota \oplus x = x$ --- that is, the $\oplus$-identity element.

Examples: $\max() = -\infty$, $\min() = \infty$, $+() = 0$, $\gcd() = 0$, etc.

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 I don't think the definition of $0!$ is motivated by the Gamma function... – Mariano Suárez-Alvarez♦ Mar 8 '11 at 17:59 @Mariano: Usually it's the reverse when talking about motivation. But there's no reason why you couldn't define factorial as a special case, since Gamma can be defined concretely in various other ways. – Quadrescence Mar 8 '11 at 18:03 Well, you could define factorials as the inverse of the Taylor coefficients of the exponential function if you really insist... :) – Mariano Suárez-Alvarez♦ Mar 8 '11 at 18:11 @Mariano: Hey, defining things as a special case is more common than defining things in a contrived manner. – Quadrescence Mar 8 '11 at 18:22

Building on Qiaochu's answer, set theory defines functions from $X$ to $Y$ as particular subsets of $X \times Y$ : a subset $f \subset X \times Y$ is a function if $\forall x \in X \exists ! y \in Y, (x,y) \in f$.

If you use this definition, then there is exactly one function from $\emptyset$ to $\emptyset$, which is $\emptyset$ (and it's also a bijection !), so there is exactly one permutation of the empty set.

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