Factorial of 0 - a convenience? [duplicate]

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.