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When doing maths that involve factorials I am often not sure as to how I should present the. i.e. $$ n! = n \times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1. $$ But what if $n$ was 3? Then you'd be squaring 3, 2, and 1. So is it best, in such a situation, to just write it as $n!= n \times (n-1) \times (n-2) \times \cdots$ ? Or does it not matter? What does "..." really mean?

Thanks.

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    $\begingroup$ Wanna avoid dots ? Write it as: $n!=\prod_{i=1}^n i$ or maybe as: $n!=\int_0^{\infty} t^n e^{-t} dt$ $\endgroup$ – Renato Faraone Oct 30 '15 at 15:40
  • $\begingroup$ I think the dots are not rigorous notation, and your equation is not really a definition, but rather a graphic representation of an idea that leads to a definition. It's used when the author considers that the definition is pretty straightforward from the graphic idea. In your case, it should be "obvious" that you should not square $3, 2$ or $1$ when starting from $3$. I personally like dots, but it's just a matter of taste. $\endgroup$ – dafinguzman Oct 30 '15 at 16:22
  • $\begingroup$ Thanks for the response. Appreciate it. $\endgroup$ – TabulaSmaragdina Oct 30 '15 at 23:33
  • $\begingroup$ If you think that someone might really multiply twice by $3$ and $2$ when $n$ is small, you might write something like $n! = n \times (n-1) \times (n-2) \times \cdots \times 1$. Not quite as clear, I tend to think, although the likelihood of misunderstanding seems small in either case. $\endgroup$ – Brian Tung Oct 30 '15 at 23:33
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First, that original representation you have should not have $(n-2)!$, but just $(n-2)$.

The "..." is just to show you are continuously multiplying by the next lowest integer until you reach $1$. For someone who has never learned about the factorial that second notation could be confusing (it may seem obvious to most to stop at $1$, and not $0$ or negative numbers, but it's good to be clear in the beginning).

For the case $n=3$ as you asked, you wouldn't list $3$ twice in the product, one of the $(n-something)$ would be your $3$, and you would multiply all lesser integers (one time each) until you reach $1$ (which means you would just multiply $3$, $2$ and $1$).

So $3! = 3 \times 2 \times 1$ is all you need and when you see

$n! = n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1$

it just means take your $n$ and multiply it with all the smaller integers less than $n$. The "..." is just a placeholder for all the other integers less than $n$. For $n=3$ your $n$, $(n-1)$, and $(n-2)$ are already your $3$, $2$, and $1$ values, so you don't list them twice.

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  • $\begingroup$ @TabulaSmaragdina The right way to thank someone for an answer on this site is to accept it - click the checkmark. You can also vote it up. In fact you can vote up several answers if several are hellpful even if you only accept one. You should also upvote other questions and answers you learn from. $\endgroup$ – Ethan Bolker Oct 30 '15 at 23:35
  • $\begingroup$ @TabulaSmaragdina, You are welcome, my pleasure. $\endgroup$ – CSCFCEM Nov 2 '15 at 13:48

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