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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: And here is another one:… (edit: second is mentioned by Moron) – Eric Naslund Mar 8 '11 at 18:06
up vote 5 down vote accepted

I don't see any difficulty with the idea of "arranging a set of zero things"... There is no need to


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…. 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

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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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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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

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