Number of subsets of a set $S$ of n elements , is $2^n$ I know that Number of  subsets of a set $S$ of size n ,  is given by the binomial sum 

$\sum_{k=0} ^n \binom{n}{k}=2^n$ ,  n=1,2,3,..   elements 

how we could conclude of prove this formula ? how to prove that this gives the number of subsets . where did this formula come from ? 
PS:- i know how to prove $\sum_{k=0} ^n \binom{n}{k}=2^n$
 A: Before I address your question on the binomial formula, here is an easier (and, IMHO, more entertaining) proof for the number of subsets of a finite set. Suppose that $S$ is a set with $n$ elements, where $n$ is a non-negative integer. To fix ideas, suppose that, say, $n=5$. Write down the elements next to each other:
$$\begin{array}{ccccc}x_1&x_2&x_3&x_4&x_5\end{array}$$
Consider the subset $\{x_2,x_4,x_5\}$. This subset can be represented as follows:
$$\begin{array}{ccccc}x_1&x_2&x_3&x_4&x_5\\\times&\checkmark&\times&\checkmark&\checkmark\end{array}$$
Here, $\checkmark$ denotes that you include a given element in the subset and $\times$ denotes that you leave it out. In an analogous fashion, every subset of $S$ can be uniquely represented as a collection of $\checkmark$'s and $\times$'s. Another example:
$$\begin{array}{ccccc}x_1&x_2&x_3&x_4&x_5\\\times&\times&\times&\times&\times\end{array}$$
represents the empty set. (Right?)
Given this representation of subsets by labeling the elements with $\checkmark$'s and $\times$'s, the number of subsets of $S$ is given by the number of ways in which $\checkmark$'s and $\times$'s can be assigned to the elements of $S$. To each of the $n$ elements, you can assign $2$ labels ($\checkmark$ or $\times$), so the total number of ways in which $\checkmark$'s and $\times$'s can be assigned, and hence the total number of subsets, is $2^n$.

As for the proof based on the binomial formula, consider first the zero-element subsets of $S$. Clearly, there is only $1=\binom{n}{0}$ such set (the empty set). The number of one-element subsets is $n=\binom{n}{1}$. The number of two-element subsets is $\binom{n}{2}$. In general, for any $k\in\{0,1,\ldots,n-1,n\}$, the number of $k$-element subsets is $\binom{n}{k}$, because this is the number of ways you can pick $k$ elements out of $n$ without repetition. It follows that the total number of subsets is the number of zero-element subsets plus the number of one-element subsets and so on. Hence,
$$\sum_{k=0}^{n}\binom{n}{k}=2^n$$
is the number of all subsets of $S$.
A: One way I like to think about this is as $(1+1)^n=\sum_{k=0}^n\binom{n}{k}$ by the binomial theorem.  When you expand out $(1+1)^n=(1+1)(1+1)\cdots$, you can interpret the $i$th term as "either the $i$th element is in the subset or it isn't."
More formally, define $F(n)$ to be the number of subsets of a set of size $n$, with $F(0)=1$ (i.e., only the empty set is a subset).  Then, it should be clear that $F(n)=2F(n-1)$ for $n\geq1$, since we can construct the subsets of a set of size $n$ by taking all of the subsets of the first $n-1$ elements, and then taking all of the subsets of the first $n-1$ elements, adding in the $n$th element.  One can show that $F(n)=2^n$, and regarding this as $(1+1)^n$, again, one can apply the binomial theorem.
