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The caption says the following:

If $n$ is an integer such that $n \ge 0$ then $n$ factorial is defined as,

$$n!=n(n-1)(n-2)\cdots(3)(2)(1)$$

if $n \ge 1$ by definition.

I'm really just confused by the $(3)(2)(1)$ in the formula, and if $n!$ is really just (for example): $5! = (5)(4)(3)(2)(1)$ then why wouldn't this just end when it hits $(n-4)$ if $n$ is 5?

I'm trying to understand this since sequence and series requires the cancellation of factorials under certain tests.

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  • $\begingroup$ the $(3)(2)(1)$ just implies it will continue until it reaches $1$. so $5!$ will stop at $n-4$ which is $1$ $\endgroup$ – kaine Nov 21 '13 at 16:10
  • $\begingroup$ "Why wouldn't this just end when it hits $(n-4)$ if $n$ is $5$" ... it does end with $(n-4)$ when $n=5$. But $n!$ isn't just defined for $n=5$, is it? $\endgroup$ – anon Nov 21 '13 at 17:17
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It seems that your confusion here is with the presentation of the formula; so, let me give you a slightly different description. Perhaps that description will make the notation you used make more sense.

For $n\in\mathbb{N}$, we define $n!$ to be the product of all natural numbers between $1$ and $n$, inclusive: that is, $$ n!=\prod_{i=1}^{n}i=1\cdot2\cdot3\cdots\cdot(n-1)\cdot n. $$ The "$\cdots$" notation takes a bit of getting used to; it is basically intended to mean "and so on" or "continuing in this way". It expresses that there's a pattern at work, and that the pattern continues on in the obvious way. The first few terms are used to establish the pattern, and the final terms describe where the pattern stops.

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  • $\begingroup$ 17 seconds earlier than mine. Wow. +1 for the nice way of stating it. $\endgroup$ – Ahaan S. Rungta Nov 21 '13 at 16:13
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The final (1) is just for completeness. Ending at n-(n-1) would make no difference.

The factorial is just defined as the product of the first N strictly positive integers up to N.

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Essentially the same answer as others have given but a slightly different way to look at it.

$n!$ is only defined for non negative integers and $n! = 1$ if $n=0$ or $n = 1$, $n! = n \cdot (n-1)!$ otherwise.

So for example $5!$

$$ $$\begin{align} 5! & = 5 \times 4! \\ & = 5 \times 4 \times 3! \\ & = 5 \times 4 \times 3 \times 2! \\ & = 5 \times 4 \times 3 \times 2 \times 1! \\ & = 5 \times 4 \times 3 \times 2 \times 1 \\ & = 120 \end{align}$$ $$

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I think the OP's problem here might be with the way the "$\dots$" notation is used in mathematics, to indicate a sequence (in this case the sequence of factors to be multiplied) by listing the first few terms and (if the sequence is finite) the last few terms, leaving it to the reader to mentally fill in the rest (on the assumption that the desired pattern will be clear). When this notation is used for a sequence that might be fairly long (for large $n$ in the case of $n!$) but might also be quite short (for small $n$), the result is that the explicitly listed first few terms and last few terms might, in the case of short sequences, be more than the terms actually intended. The rule for understanding this notation is to figure out the pattern and then interpret it, even in the case of short sequences, as beginning and ending as indicated but, in between, following the pattern, not necessarily including all the terms that were written to show the pattern.

Thus, in the case of $n(n-1)(n-2)\cdots3\cdot2\cdot1$, the pattern is "start with $n$ and count down (decreasing the factor by 1 at each step) until you reach 1". For any $n\leq5$, that will involve fewer terms than what is written. For example, if you take what is written literally for $n=4$, you'd get $4\cdot3\cdot2\cdots3\cdot2\cdot1$, but the correct interpretation does not repeat the $3$ and the $2$. When $n=2$, the literal reading is even worse, since the factor $(n-2)$ is then $0$, but the intended meaning is still "start at $n$ and continue to $1$", so you get $2\cdot 1$ (not $2\cdot1\cdot0\cdots3\cdot2\cdot1$).

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Going down to $ 3, 2, 1 $ is saying that it will keep multiplying until it reaches $1$, which is basically saying it will keep multiplying until it reaches $ n - \left( n - 1 \right) = 1 $.

What you are saying is, for $n=5$, it goes till $n-4$. In general, it goes till $n-(n-1)$. But this is just $1$. It's equivalent. Concisely, it is $$ n! = \displaystyle\prod_{k=1}^{n} k. $$

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