factorial

Factorial

$$n!:=\prod _{i=1}^{n}i=\{\begin{array}{ll}1& \text{if }n=0\\ n\cdot (n-1)!& \text{if }n>0\end{array}$$

Simplifications

$$\begin{array}{r}\frac{n!}{(n-k)!}=n\cdot (n-1)\cdots (n-k+1)\end{array}$$

… count down $k$ terms from $n$.
Only the first $k$ terms of $n!$ remain after cancellation, the last $(n-k)!$ terms cancel out.
E.g.:

$$\frac{5!}{(5-2)!}=\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1}=5\cdot 4=20$$

Why is $0!=1$?

$$\begin{array}{rlr}4!& =4\cdot 3\cdot 2\cdot 1=24& ÷4\\ 3!& =3\cdot 2\cdot 1=6& ÷3\\ 2!& =2\cdot 1=2& ÷2\\ 1!& =1& ÷1\\ 0!& =1& \end{array}$$

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There’s only one way to arrange a group of zero objects: do nothing.

The factorial tells us how many ways we can arrange $n$ objects

permutations: How many ways are there to arrange all elements of $A$?

$\pi :A\to A$, bijective permutation function
$f:x_1,x_2,\dots x_n\mapsto (x_1,\dots ,x_n),x_i=\pi (a_i)$
$f:S_A\to \{x\in A^n:\forall i\ne j:x_{i\ne x_j}\}$
$|S_A|=\#(x_1,\dots ,x_n):\forall i\ne j:x_i\ne x_j$ (exactly what we had in (ii) above)
$=n!\left(\genfrac{}{}{0ex}{}{n}{n}\right)=n!$
→ $|A!|=n!$

Link to original

$n!$ = number of ways to visit n points exactly once

At each step, you have one fewer choice, giving you $n\cdot (n-1)\cdot (n-2)\cdots 1$ options: interactive visualization