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In the formula,

$$\frac{n(n-1)(n-2)\cdots(n-r+2)}{(r-1)!}a^{n-r+1}b^{r-1}$$

what does the "$\cdots$" mean?

3 Answers 3

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It means: "There are too many terms to write, but follow the obvious pattern to fill them in".

In your example, you subtract $1$ from a factor to get the next factor. I might read that aloud as "$n$ times $n-1$ times $n-2$ all the way down to $n-r+2$".

As another example, $$ 3 + 6 + 9 + \cdots + 3n $$ would indicate the sum of all positive multiples of $3$ less than or equal to $3n$.

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    +1 Good answer. Simple. Easy to understand. Does this mean that I should solve the $(n-r+2)$ first?2012-09-04
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    @DantheMan Probably the $r$ in this case is meant to be some fixed constant. For example, if $r = 5$, then your formula simplifies to $$\frac{n(n-1)(n-2)(n-3)}{4!}a^{n-4}b^4.$$ I can't really say more without context.2012-09-04
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    Actually, this is the formula for finding the $r$th term of a binomial $(a+b)^r$2012-09-04
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    Ah right.. the $r$th term of $(a+b)^n$. What I mean by that is, should I solve $(n-r+2)$ first, so that I know when the obvious pattern of $\cdots$ ends?2012-09-04
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    @DantheMan There are (at least) two ways to think about it. One is what you are suggesting: "Since $r = 5$, the last number in my pattern is a 3." Another way is to notice that there are always $r-1$ factors in the product. So you might think: "Since $r = 5$, I keep following the pattern until I've written four factors down."2012-09-04
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    Awesome. Thanks.2012-09-04
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    @Dan Not only may there be too many terms, but, the number of terms may not be fixed, e.g. $\,r-1\,$ terms in your product2012-09-04
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Writing it as "$\cdots\;$", would be better than "$\dots\;$" . It indicates a product:

$$ \frac{n(n-1)(n-2) \cdots (n-r+2)}{(r-1)!}a^{n-r+1}b^{r-1} =\frac{\prod_{k=0}^{r-2} (n-k)}{(r-1)!}a^{n-r+1}b^{r-1} $$

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    This would be better as a comment.2012-09-04
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    @Chaz, doesn't it answer the question?2012-09-04
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    For those of us who *already know the answer*, it may be sufficient. But for someone who doesn't know...? Either way, no big deal2012-09-04
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    Also, I'm fairly certain that there was a ninja edit to this answer2012-09-04
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    ...as there was to the question.2012-09-04
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    At first it only said, Writing it as "$\cdots\;$", would be better than "$\dots\;$".2012-09-04
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    That's probably why @TheChaz said that.2012-09-04
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    I thought about giving it more detail, BEFORE I read the comments. Is that wrong? "Either way, no big deal "2012-09-04
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    Ah yeah. No problem.2012-09-04
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It means multiplying a set of terms ($n$, $n-1$, $n-2$, $\dots$(!), $n-(r-2)$) when each term is the result of subtracting $0$, $1$, $2$, $\dots$, $r-2$ from $n$, respectively.