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?
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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 $3n$.
Austin Mohr
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1+1 Good answer. Simple. Easy to understand. Does this mean that I should solve the $(n-r+2)$ first? – Daniel Sep 04 '12 at 13:02
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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. – Austin Mohr Sep 04 '12 at 13:05
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Actually, this is the formula for finding the $r$th term of a binomial $(a+b)^r$ – Daniel Sep 04 '12 at 13:10
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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? – Daniel Sep 04 '12 at 13:19
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1@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." – Austin Mohr Sep 04 '12 at 13:25
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Awesome. Thanks. – Daniel Sep 04 '12 at 13:34
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1@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 product – Bill Dubuque Sep 04 '12 at 13:55
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Writing it as "$\cdots\;$", would be better than "$\dots\;$" . It indicates a product:
$$ \frac{n(n-1)(n-2) \cdots (n-r+2)}{(r-1)!}a^{n-r+1}b^{r-1} =\frac{\prod_{k=0}^{r-2} (n-k)}{(r-1)!}a^{n-r+1}b^{r-1} $$
draks ...
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1For those of us who already know the answer, it may be sufficient. But for someone who doesn't know...? Either way, no big deal – The Chaz 2.0 Sep 04 '12 at 14:12
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1Also, I'm fairly certain that there was a ninja edit to this answer – The Chaz 2.0 Sep 04 '12 at 14:14
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2At first it only said, Writing it as "$\cdots;$", would be better than "$\dots;$". – Daniel Sep 04 '12 at 14:17
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I thought about giving it more detail, BEFORE I read the comments. Is that wrong? "Either way, no big deal " – draks ... Sep 04 '12 at 14:19
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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.
Gigili
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