0!=1

AfricanBorn · #1 $Old$ November 8th, 2009, 03:04 AM

I've got a very simple question yet it has been bugging me for ages, WHY/HOW does 0 permutations/arrangements =1 (0!=1)?

Prove It · #2 $Old$ November 8th, 2009, 03:33 AM

It's simply a matter of definition - and it is a useful definition at that.

But if you wanted to be really clever, note that

$n! = n(n - 1)!$

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

By letting $n = 1$ we have

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

$0! = 1$ .

Laurent · #3 $Old$ November 8th, 2009, 03:55 AM

Quote:

Originally Posted by AfricanBorn $View Post$

I've got a very simple question yet it has been bugging me for ages, WHY/HOW does 0 permutations/arrangements =1 (0!=1)?

$n!$ is the number of permutations of the set $\{1,\ldots,n\}$ , i.e. the number of maps from $\{1,\ldots,n\}$ to itself that are bijective.

If $n=0$ , then $\{1,\ldots,n\}=\emptyset$ , and there exists one map from $\emptyset$ to $\emptyset$ ... the empty map $\emptyset$ !!

Remember how maps are defined: a map from $A$ to $B$ is a subset $f$ of $A\times B$ such that, for every $x\in A$ , there exists exactly one $y\in B$ such that $(x,y)\in f$ , and we write $f(x)=y$ .

If $A=\emptyset$ , then $f=\emptyset$ is a subset of $A\times B(=\emptyset)$ (whatever $B$ is) that satisfies the assumption since there is no $x\in A$ (hence the condition is automatically fulfilled: it is empty). Thus, $\emptyset$ is a map from $\emptyset$ to anything. And it is the only one.

It is bijective from $\emptyset$ to $\emptyset$ : for every $y\in\emptyset$ , there is a unique $x\in\emptyset$ mapped to $y$ . Indeed, there is no such $y$ , so there's nothing to be checked.

Thus, $\emptyset$ is the only map from $\emptyset$ to itself, and it is bijective. As a conclusion, $0!=1$ .

This makes sense in many ways, like Prove It illustrated.