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Proving a special case of the binomial theorem

Prove the identity using a combinatorial argument: $$\sum_{k=0}^{n}\binom{n}{k} = 2^{n}$$

I'm not sure how to do a combinatorial argument so any help is appreciated. All I know is that $2^{n}$ is representing the number of elements in the power set of n.

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    like [this](http://math.stackexchange.com/questions/67889/why-does-sum-limits-i-0k-k-choose-i-2k/67895#67895) or [this](http://math.stackexchange.com/questions/27539/proving-a-special-case-of-the-binomial-theorem)?2012-08-01
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    @Jason Curt: Please why are you asking two similar question differently? You have the chance to marge the two questions together. Take a look at this (http://math.stackexchange.com/questions/177405/prove-by-induction-2n-cn-0-cn-1-cdots-cn-n ).2012-08-01

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