How does factorial result in the computation of possible orderings?

Question

For 3 characters to find the ways of putting them in order: ABC, ACB, BAC, BCA, CAB, CBA $= 3! = 3\times2\times1 = 6$

When trying to find ways to put an object in order $n\cdot(n-1)\cdot(n-2)\cdots2\cdot1=n!$, what is happening upon this factorial computation with an usage of multiplication?

I can't wrap my head around the reason factorial can compute this and trying to understand its underlying theory.

Answer 1

For a set $S$ of items, let $A(S)$ denote the number of ways to arrange the items in $S$ in sequence. If you have a set $\{1,\dotsc,n\}$ of $n$ items, you have $n$ possible choices for the item to place first, and each choice leaves a different set of items for the remaining places, so you have

$$A(\{1,\dotsc,n\}) = \sum_{k=1}^n A(\{1,\dotsc,n\}\setminus\{k\}).$$

But the number of ways to arrange the items depends only on the number of items, and not on which items they are, so if we call the number of ways to arrange $m$ items $B(m)$, then we have

$$B(n) = \sum_{k=1}^n B(n-1) = n\cdot B(n-1),$$

and that leads to see that $B(n) = n!$.

Answer 2

The basic idea is that of multiplication. If you have $n$ choices for one object and for each of them you have $m$ choices for a second, you have $nm$ choices for the pair. Then recognize that in putting $n$ items in order, after you chose one of $n$ for the first item, you have $n-1$ choices for the next because the first isn't available any more. Then you have $n-2$ choices for the third because two items have been removed and so on.