I understand that factorial represents the total number of possible ways to arrange $n$ number of items and it is calculated as $n\cdot(n-1)...\cdot2\cdot1$ . I don't understand though how this formula can be derived by using just the knowledge of multiplication.
Consider the example of having 4 chocolates $A, B, C$ and $D$ and finding the number of possible ways of arranging them together. The answer for this is $4\cdot3\cdot2\cdot1=24$ but I don't understand in the context of this example, what $4\cdot3$ actually means. Does it mean that there are 4 possible candidates out of $A, B, C$ and $D$ for the first position and for the second position, there are 3 possible candidates so with one out of those 3 candidates matches with the 4 possible candidates for the first position and forms $4$ arrangements. So if you can form 4 arrangements with 1 of 3 candidates, then you can form $4\cdot3=12$ arrangements with 2 out of 4 chocolates. But the idea of forming 4 arrangements with a single candidate for the second position does not seem correct because one of those four arrangements will be a candidate with itself ($AA$ for example) and that makes no sense. So this is the wrong way to think about it.
So what is the correct way to think about the meaning of $4\cdot3$ in this context? Do we say that 1 out of 4 candidates for the first position goes along with 3 candidates for the second position, so with 2 out of 4 chocolates, we can form $3\cdot 4 = 12$ candidates? How do we proceed from here to the step of $(3\cdot4)\cdot2$? What does $(12)\cdot2$ practically mean? Does it mean we have possible candidates to pick from and with a single candidate we can have 12 arrangements of 3 chocolates and with 2 candidates, we can have $12 \cdot 2 = 24$ arrangements of 3 chocolates?
I am confused.
You just have to think in terms of you can't select any candidate again. That is, once you have selected $A$, you can't select it again.
Now consider $N$ candidates for $N$ sequnetial positions, so that order matters, i.e. $AB$ is different from $BA$.
You can choose any one of the $N$ for the first position.
But for the second position, you have $N-1$ choices - you can't reuse the candidate you used for the first position. That is, for every choice of the first position, you have $N-1$ choices for the second one. This is where the multiplication principle comes in.
Take one example: consider $N=3$ with $\{A, B, C\}$.
If the choice for first position is $A$, you have 2 choices for second position: $B, C$.
If the choice for first position is $B$, you have 2 choices for second position: $A, C$.
If the choice for first position is $C$, you have 2 choices for second position: $A, B$.
Total $3 \times 2 = 6$ choices.
Similarly for the $k$th position you have $N-k+1$ choices.
The product of all is $N!$
If you consider $N$ candidates but only $r \leq N$ positions, by similar reasoning you arrive at
the formula of $\hspace{2pt}$ $^N$P$_{_{\LARGE r}}$.