How's it possible to arrange 0 objects? How can 0! = 1?

Combinatorics is not necessarily tangible, so the question of what it means to physically rearrange objects is irrelevant. On the other hand, it's easy to write down the permutations of small finite numbers of objects:

0: []
1: [1]
2: [1,2] [2,1]
3: [1,2,3] [1,3,2] [2,1,3] [2,3,1] [3,1,2] [3,2,1]

posted almost 3 years ago

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Peter Taylor‭

1200 reputation 8 43 135 42

This explanation may work for some but not for others--it is perhaps a matter of taste or intuition whether you feel that it is sensible. Is doing nothing a "way of arranging" zero objects? Eh, you can get very philosophical about this and probably not in a productive way.

I find the following argument more satisfying for why we should take 0!=1. It is because 0! is itself not meaningful, so in a sense, we should not say that it equals anything. In another sense, we are free to say that it equals whatever we want. After all, we may define any symbol to be the name of any function. So we can choose to take n! to name any function we want.

For n>0 we take n! to be the number of arrangements of n objects. We are not compelled to define n at 0. But if we want to, here's a good reason to define it to be 1: It makes writing other sensible equations easier. For instance, take the sequence

$$ 1, x, \frac{x^2}{2}, \frac{x^3}{3!}, \frac{x^4}{4!}, ... $$

In this sequence we keep increasing the degree of the monomial, and the factorial in the denominator. This is an important sequence because we can use it to define $e^x$. Notice that every term except the first can be described as

$$ \frac{x^n}{n!} $$

If we define $0!=1$ then this equation also works for the first, because then $\frac{x^0}{0!}=1$.

Here's another example: Combinations. Suppose that you want to count the number of ways to choose $k$ objects from a collection of $n$ objects. It turns out that a formula for this is given by

$$ \binom n k = \frac{n!}{k!(n-k)!} $$

and it makes sense for all $0<k<n$. But what if you want to choose zero objects from $n$ objects? Well there is just one way to do that: Choose no objects!

I think choosing no objects makes more sense than arranging zero objects, because even when you choose no objects from say 10 objects, those 10 objects are actually there. So it makes sense to me that there is one way to not choose any of the objects that are available to you: Just stand there like an idiot.

So hopefully this argument is convincing that we should have $\binom n 0 = 1$. That is to say, for any appropriate $n$, the number of ways to not choose any is 1. Well, if you take $0!=1$ then the same formula from before works. Because with that formula you get

$$ \binom n 0 = \frac{n!}{0!(n-0)!}=\frac{n!}{n!}=1 $$

and in fact it also makes sense when $k=n$.

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

That is also, intuitively the number of ways of selecting all the objects.

Many other examples are possible, where our formulas are simpler to write and continue to make sense, when we take $0!=1$.

posted almost 3 years ago

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3y ago

whybecause‭

384 reputation 6 17 46 16

How is it possible to arrange $0$ objects? Well, by doing nothing. Doing nothing is something you can do, and you can do it in exactly one way.

Another way to see it is to consider that it is also the number of ways you can rearrange the objects in a row. One way to rearrange the objects is, again, doing nothing. If there is only one object, doing nothing is indeed the only way to rearrange the objects, therefore $1!=1$. And if there are no objects at all, again, doing nothing is your only option, thus also $0!=1$.

This is different to the situation when dividing by zero. If you distribute $m$ cakes to $n$ children so that no cake is left, each child gets $m/n$ cakes. Now what happens if $n=0$? Well, if $m$ is nonzero, it is simply not possible: If you have no children to distribute your cake to, you simply won't get rid of your cake, as there's no one there to take it. On the other hand, if $m=0$ as well (you have no cake) you can distribute your zero cakes to the zero children; however you can't tell how many cakes each child will get, as any number of cakes per child will work.

Also, generally $n!$ is not defined to be the number of arrangements, but simply as the product $1\cdot 2\cdot 3 \cdots n$. It just turns out that this is exactly the number of arrangements of $n$ objects in a row.

Now with that definition, how do we see that $0!=1$? Well, clearly $0!$ is the empty product, which by definition is $1$.

Alternatively you can observe that obviously for $n>0$, $(n+1)! = (n+1)n!$. Therefore the obvious definition of $0!$ is so that this relation still holds, that is $1! = 1\cdot 0!$. Given that $1!=1$, it immediately follows that $0!=1$.

posted almost 3 years ago

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celtschk‭

1011 reputation 17 26 135 29