# Proof

(related to Proposition: Factorial)

Let $$V$$ be a finite set with $$|V|=n$$.

If $$n \ge 1$$, then without loss of generality let $$V=\{a_1,a_2,\ldots,a_n\}$$. We want to count the different permutations of $$V$$, which according to their definition are bijective maps on the index set $$I=\{1,2,\ldots,n\}$$: $$\pi:I\mapsto I$$. More precisely, instead of counting different bijective maps, we will find the cardinal number of the set $$\Pi=\{\pi:I\mapsto I\}$$ of all possible bijective maps.

In order to define one element of $$\pi\in\Pi$$, we have do define $$\pi$$ for all $$i\in I$$, using the following procedure:

To define $$\pi(1)=i_1$$, all values from $$I$$ are possible, and we choose $$i_1\in I_1:=I$$.

We have to define $$\pi(2)=i_2$$ in such a way that $$\pi$$ remains bijective. For that, only the values from $$I_2:=I\setminus\{i_1\}$$ are possible, and we choose $$i_2\in I_2$$.

Similarly, to define $$\pi(3)=i_3$$, only the values from $$I_3:=I\setminus\{i_1,i_2\}$$ are possible, and we choose $$i_3\in I_3$$.

Similarly, to define $$\pi(4)=i_4$$, only the values of $$I_4:=I\setminus\{i_1,i_2,i_3\}$$ are possible, and we choose $$i_4\in I_4$$.

$$\vdots$$

We repeat this procedure until we define $$\pi(n)=i_n$$, for which only one value from $$I_n:=I\setminus\{i_1,i_2,i_3,\ldots,i_{n-1}\}$$ is possible, and we choose $$i_n\in I_n$$.

Following the definition of set difference and the fundamental counting principle we have that

$\begin{array}{ccl} |\Pi|&=&|I_1|\cdot|I_2|\cdot|I_3|\cdot\ldots\cdot|I_{n-1}|\cdot|I_n|\\ &=&n\cdot(n-1)\cdot(n-2)\cdot\ldots\cdot 2\cdot 1\\ &=&n! \end{array}$

If $$n = 0$$, then $$V$$ is empty and the existence of only one bijective map $$\pi:\emptyset\mapsto\emptyset$$ is vacuously true. Therefore, we set

$0!:=1.$

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### References

#### Bibliography

1. Aigner, Martin: "Diskrete Mathematik", vieweg studium, 1993