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I'm reading a book that claims that $\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n}=2^{n}$

Is there a easy proof for this?

Postskjerm
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    The binomial theorem applied to $(1+1)^n$ is the most classic proof. – Crostul Apr 28 '17 at 09:00
  • @Arnaud: There's the combinatorics tag, it shouldn't be a duplicate. No, I guess it is, since it seems the OP wasn't really loooing for a combinatorial proof – Vincenzo Oliva Apr 28 '17 at 09:12
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    I don't think it's a duplicate of the question linked above (since that question is asking for a non-standard proof), but it is an exact duplicate of https://math.stackexchange.com/questions/144256/sum-from-0-to-n-of-n-choose-i?noredirect=1&lq=1 – Especially Lime Apr 28 '17 at 10:48
  • @EspeciallyLime This question does not specifiy what kind of proof is required, so I don't see why it wouldn't be a duplicate of the linked question. Besides, the link you give is itself closed as duplicate of the above linked question. – Arnaud D. Apr 28 '17 at 11:59
  • Yes it does: it specifies "easy". By contrast, the linked question is someone who already knows the easy proofs asking for something different. None of the answers below would be what the linked question was looking for, and the accepted answer there would not be the most appropriate for this question. – Especially Lime Apr 28 '17 at 14:35

4 Answers4

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We can use the binomial theorem, which states that $$ (x+y)^n = {n \choose 0}x^n y^0 + {n \choose 1}x^{n-1}y^1 + {n \choose 2}x^{n-2}y^2 + \cdots + {n \choose n-1}x^1 y^{n-1} + {n \choose n}x^0 y^n. $$ Set $x=1$ and $y=1$ to obtain the result.

Cm7F7Bb
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$$(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+........^nC_n(x)^n\tag{Binomial Expansion}$$ Just put $x=1$ and Tada!

The Dead Legend
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Say you have a set of objects of two types, say red and blue balls, otherwise indistinguishable. In how many ways can you take $n$ of them and arrange them? The classic answer is your RHS, $2^n $. But you can also think of it like this: there are $\binom{n}{0}$ ways to have $0$ red balls in our collection, $\binom{n}{1}$ ways to have $1$ red ball in it, ... , $\binom{n}{n}$ ways to have $n $ red balls in it. So the answer is the sum of all of these, i.e. your LHS, too, hence the equality holds.

Vincenzo Oliva
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There are $2^{n}$ ordered strings of zeros and ones. The number of such strings with $k$ ones is $\binom{n}{k}$. Since every such string has some number, $0 \leq k \leq n$, of ones, it must be $2^{n} = \sum_{k = 0}^{n}\binom{n}{k}$.

Dan Fox
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