prove that$\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n}=2^{n}$

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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 $2^{n} = \sum_{k = 0}^{n}\binom{n}{k}$.