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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.

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    $\begingroup$ Perhaps easier to consider the following (relatively) equivalent question: In how many ways can the elements of the set {A,B,C,D} be selected, sampling 4 characters without replacement, where order of selection is deemed relevant? Answer: There are $~4~$ choices for which of the letters will be the first one chosen. Then, regardless of which letter is the first one chosen, there will be exactly $~3~$ choices for which of the remaining letters is the second one chosen. ...see my next comment $\endgroup$ May 20, 2023 at 4:33
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    $\begingroup$ You seem to be thinking of it backwards. For each candidate in the first position, you want a candidate in the second position. There are only three possible choices for second position (i.e. the ones that weren't chosen first), so each choice for the first position yields three possible choices for the second position - something like $AA$ isn't possible because you are avoiding it by your choice in the second position. $\endgroup$ May 20, 2023 at 4:35
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    $\begingroup$ Then, regardless of which two letters are the first two letters chosen, there will be exactly $~2~$ choices for which of the remaining letters is the third one chosen. Then, regardless of which three letters are the first three chosen, there will be exactly one choice for the last letter chosen. So the number of permutations equals $~4 \times 3 \times 2 \times 1.~$ This is a consequence of the fact that you multiply the factors that represent the number of available choices for each character position. $\endgroup$ May 20, 2023 at 4:36
  • $\begingroup$ My answer to the question below is a good simpler example. Never mind the downvote... It all has to do with the tree structure:math.stackexchange.com/questions/4701521/… $\endgroup$
    – NoChance
    May 20, 2023 at 4:40
  • $\begingroup$ @MandelBroccoli There are only three possible choices for second position (i.e. the ones that weren't chosen first), so each choice for the first position yields three possible choices for the second position - something like AA isn't possible because you are avoiding it by your choice in the second position. Ok makes sense. What is the intuitive explanation for the next part (3*4)*2? Is it that (3*4) arrangements can go along with 1 candidate, so how many arrangements can you form with 2 of them? $\endgroup$
    – a_sid
    May 20, 2023 at 14:51

1 Answer 1

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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$_{_{\LARGE r}}$.

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  • $\begingroup$ thank you for the response $\endgroup$
    – a_sid
    Nov 14, 2023 at 6:28

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