How many ways can you arrange nothing?

Question

I have heard that you can arrange nothing in only $~0! = 1$ way. That is, leave it like that, nothing. Is that reasoning correct? I want to understand why there's one way to arrange nothing when there's nothing to be arranged in the first place. You can't arrange something that doesn't exist, right? I hope you help me understand this concept.

See also this post. "Nothing" means here choosing $0$ items. And it is indeed possible to choose 0 items from $n$ items. — Dietrich Burde, Jun 16 '22 at 08:32
The technically "correct" reasoning is that $0! = 1$ by the very definition of the factorial function. However, remembering that zero items can be arranged one way because there's nothing to arrange is a good mnemonic for remembering the value of $0!$. — Theo Bendit, Jun 16 '22 at 08:34
And we had the discussion why $0!=1$ is the "correct" definition already here. — Dietrich Burde, Jun 16 '22 at 08:36
Nothing is not mathematical term. Arrangement of some set is equivalent to sequence containing all elements of this set which contains each element exactly one time. For empty set there is exactly one such sequence (empty sequence), that's why there is only one arrangement for set containing 0 elements. — Ivan Kaznacheyeu, Jun 16 '22 at 13:50

score 10 · Answer 1 · answered Jun 16 '22 at 08:42

First, some conceptual stuff before we come to the actual question. There is an important conceptual difference between the "number of ways to arrange things" and the "number of arrangement", at least in everyday language. The former tells us how many ways there are to do something, while the latter tells us how many ways there for things to be. If you think in the former way, you will (rightly) say that we don't have to actually do anything to "no objects", so there is no way to arrange them. But in math, we have for the most parts abandoned this so called dynamic way of thinking in favor of the latter, static way of thinking. We don't actually count ways to do things. We count the finished states. And here comes a language barrier: mathematicians sometimes still talk in a way that sounds like they think dynamically, but they actually mean it in a static way. So when you hear "number of ways to arrange things" in a mathematical context, what you should take it to mean is "number of finished arrangements".

Now with this conceptual bit out of the way, here is an intuitive explanation why there is one arrangement of nothing:

Imagine a family of $n$ people, seated at a table. The table is surrounded by $n$ chairs. How many possible seating arrangements are there? Of course it's $n!$. And what happens if $n=0$? Then there is a table, but with no chairs and no people. But the empty table is still there, with an "empty" seating arrangement. This is the one arrangement of nothing. Basically, imagine that the space where the things would be arranged is still there, just empty.

In a rigorous formulation, we can see it this way: every arrangement is a bijective function $f:X\to Y$, where $X$ is the set of family members and $Y$ the set of chairs. Counting arrangements means counting the number of such functions. If $n=0$, both sets are empty (empty set of chairs is the empty table), so the only possible function is the empty function. That's one function/arrangement.

score 2 · Answer 2 · answered Jun 16 '22 at 08:38

2

You can see it as the number of different tuples:

For the set $\{1, 2\}$ the possible permutations are $(1,2)$ and $(2,1)$.

For the set $\{\}$ the only possible permutation is $()$

answered Jun 16 '22 at 08:38

T C Molenaar

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robjohn · Answer 3 · 2022-06-16T23:53:24.453

The number of ways to rearrange $n$ items is the number of bijections on $N=\{1,2,\dots,n\}$. That is the number of $F\subset N\times N$ so that

for all $k\in N$, there is an $m\in N$ so that $(k,m)\in F$ ($N$ is the domain of $F$)
if $(k,m_1)\in F$ and $(k,m_2)\in F$, then $m_1=m_2$ ($F$ is a function)
if $(k_1,m)\in F$ and $(k_2,m)\in F$, then $k_1=k_2$ ($F$ is an injection)
for all $m\in N$, there is an $k\in N$ so that $(k,m)\in F$ ($F$ is a surjection).

There is only $1$ bijection on the empty set, the empty function, $F=\{\}$. Therefore, the number of ways to rearrange the empty set is $1$.

score 1 · Answer 4 · answered Jun 17 '22 at 00:17

Stack High School is having tryouts for next year's basketball teams. Interested students line up at the signup table in the gym. Because there are separate boys' and girls' teams, there are two separate lines. The number of possible ways to line up the students is thus $b!g!$, where $b$ is the number of boys and $g$ is the number of girls.

The next day, they have tryouts for (American) football. Because there's not (yet) a girls' team, only boys show up. Thus there are only $b!$ possible ways to line up the students.

But it turns out that our original formula of $b!g!$ is still valid, because $g! = 0! = 1$. Had we defined $0! = 0$ instead, we'd get $b!g! = 0$ instead, which is clearly wrong, because even with no girls there, it still makes sense to ask how many ways we can arrange the boys.

And that's one motivation for saying that “there is one permutation of zero items” or $0!=1$. It may not make much intuitive sense in isolation, but it makes the math work out when there are multiple sets of items to arrange, and one of those sets could be empty.

How many ways can you arrange nothing?

4 Answers4