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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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    @Cobold: do you see the comment above by Mark Bennet? $\binom{n}{i}$ is the binomial coefficient ("n-choose-i"); for each integer i, $0\leq i \leq n$ (between 0 and n, inclusive), you need to compute $\binom{n}{i} a^{n-i}b$, and ultimately sum all such computations. $\binom{n}{i}$ is defined in Mark's comment above (he uses n-choose-r, so just substitute i for r).2011-07-02

3 Answers 3

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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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    @GEdgar Thanks!2011-12-23
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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}