Prove the identity $\sum_{k=0}^n \binom{n}{k}=2^n.$ using combinatorial proof Prove the identity $\sum_{k=0}^n \binom{n}{k}=2^n.$ using combinatorial proof.
Suppose you have $n$ distinct objects and you want to choose some (possibly none) of them. 

One way to look at this is as follows. You can either select no object, or one object, or two objects, and so on, you can select all the $n$ objects. The number of ways of selecting $k$ objects is $\binom{n}{k}$, and as $k$ varies from $0$ to $n$, the total value is $$\sum_{k=0}^n\binom{n}{k}$$

Another way is to look at each object. For every object, you have two choices - either you choose it or you don't. Thus, for $n$ objects, you have $2^n$ choices, and by varying your choices, you can select any number of objects in any configuration.

As both the above refer to counting the number of ways of doing the same thing, they must be equal. So,
$$\sum_{k=0}^n\binom{n}{k}=2^n$$
Use the binomial theorem $(1+x)^n=\sum_{k=0}^n {n\choose k}x^k$, with $x=1$.