(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
\[0!:=1.\]