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What do the parentheses next to the summation involving the binomial coefficients mean? Like this: $$\sum _{i=0}^{n} \binom{n}{i}a^{(n-i)}b^i=\left(a+b\right)^n $$

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It's hard to interpret your question without seeing what you can see. Perhaps you could use codecogs.com/latex/eqneditor.php and then post the code here, putting it in between two dollar signs -- $$. –  InterestedGuest Jul 2 '11 at 20:22
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The most obvious answer would be something to do with $\binom{n}{r} = \frac{n!}{r!(n-r)!}$. If it is, do edit it into your own question, because what you state is very unspecific. Or respond so someone else can edit it in. –  Mark Bennet Jul 2 '11 at 20:29
    
@MarkBennet: OP has edited his original post (that did not say anything about binomial coefficients) so combinations are indeed probably what he means. @Cobold: check out en.wikipedia.org/wiki/Combination this link. If you understand combinations and summations apart from each other, you can just combine them. There is nothing really special to it. Or if your question is more concrete, please rephrase. –  user12205 Jul 2 '11 at 20:34
    
@Jeroen Vaelen: I think that in most books the standard notation is that after the $\displaystyle\sum$-sign everything is summed. So $$\displaystyle\sum_{i=0}^3(2\times i)+1=\displaystyle\sum_{i=0}^3(2\times i)+\displaystyle\sum_{i=0}^3 1.$$ –  Américo Tavares Jul 2 '11 at 20:38
    
@JeroenVaelen I mean the paranteses (n k) next to the capital sigma, could you explain what does that mean. There's no dividing there, that's what I don't understand. –  Cobold Jul 2 '11 at 20:41

3 Answers 3

up vote 9 down vote accepted

They are not parentheses. It is the binomial coefficient, as Mark Bennet commented.

$$\binom{n}{i}=\dfrac{n!}{i!(n-i)!}.$$

Given $n$ objects it counts the number of ways of choosing $i$ objects from those $n$, i.e. the combinations of $n$ objects taken $i$ at a time without repetitions.

Alternative notation: $C(n,i)$ or $C_i^n$ (or $C_n^i$).

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$C_n^i$ rather than $C_i^n$. –  Did Jul 2 '11 at 21:47
    
@Didier Piau: Please excuse me. According to Wikipedia the notation $C_i^n$ is equivalent to $\binom{n}{i}$ ("n choose i"). How is $C_i^n$ read? (in Portuguese is "combinações de n, i a i"). –  Américo Tavares Jul 2 '11 at 22:23
    
@Didier Piau: I have read Wikipedia again and it may be $C_n^i$ too. I only knew $C_i^n$ with the same meaning of $\binom{n}{i}$. –  Américo Tavares Jul 2 '11 at 22:34
    
@Didier Piau: I added the alternative notation $C_n^i$. –  Américo Tavares Jul 2 '11 at 22:36
    
@Cobold: What the notation means is the following. If you have $n$ such objects to choose from, you only want to choose $i$ objects from the total $n$. For example: If you have $n$ marbles inside of a jar but you only want all of the green marbles.. then you would choose $i$ marbles from the batch of $n$ marbles. So if there were a total of 5 marbles ($n$) and two were green, and you wanted all the green's $~(i)$,then you could use the notation $n\choose i$=$5\choose 2$. Meaning, out of the $5$ marbles, I only choose $2$. Hope that this cleared up the meaning some for the binomial coefficient. –  night owl Jul 3 '11 at 3:07

The way it works if you expand $(a+b)^n$ is that there are $n$ brackets and you have to choose an $a$ or a $b$ from each one. There are 2 choices for each bracket, hence 2^n choices overall.

The sum you have given gathers all the terms with $i$ copies of $a$ and $n-i$ copies of $b$. The bracket $\binom{n}{i}$ is the number of ways of doing this, and is called n-Choose-i (which is the reason for the $C$ in some alternative notations) or a Binomial Coefficient (binomial because the original brackets contain two terms $a$ and $b$).

It might be interesting for you to see whether you can get any intuition or insight for why the value of the bracket is what it is.

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It is read "$n$ choose $i$" and is equal to the number of ways to choose $i$ objects out of $n$ different objects, where the objects are pairwise different. This number is given by the formula \begin{equation} {n \choose i}=\frac{n!}{i!(n-i)!}. \end{equation}

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