Binomial Theorem and number of subsets I'm reading my textbook on binomial theorem and there's an example question with the solution.  I don't understand it.  Can someone please explain?  Thanks in advance.
How many subsets are there of a set consisting of n elements?
Solution: $\displaystyle\sum\limits_{k=0}^n \binom{n}{k} = (1 +1)^n = 2^n$
What does this have to do with the binomial theorem and where is the $x^k  y^{n-k}$ terms?
http://en.wikipedia.org/wiki/Subset
 A: Let $S$ be a set of $n$ objects; then the binomial coefficient $\binom{n}k$ is the number of $k$-element subsets of $S$. Thus, $\sum_{k=0}^n\binom{n}k$ is the number of subsets of $S$ of all possible sizes from $0$ through $n$.
The binomial theorem says that
$$(x+y)^n=\sum_{k=0}^n\binom{n}kx^ky^{n-k}\;;\tag{1}$$
if you substitute $x=y=1$ in $(1)$, you get
$$(1+1)^n=\sum_{k=0}^n\binom{n}k1^k\cdot1^{n-k}=\sum_{k=0}^n\binom{n}k\;.\tag{2}$$
And of course $(1+1)^n=2^n$, so $(2)$ reduces to
$$2^n=\sum_{k=0}^n\binom{n}k=\text{number of subsets of }S\;.$$
A: Also, you can see another way to get the RHS of your equation. Consider a subset. There are $n$ possible members, and each can either be in the subset, or not, i.e. there are 2 possibilities for each member. 
Therefore the number of total possible subsets is $2*2*2...$ $n$ times = $2^n$.
A: How many subsets are there of a set consisting of n elements?
Solution: $\displaystyle\sum\limits_{k=0}^n \binom{n}{k} = (1 +1)^n = 2^n$
What does this have to do with the binomial theorem and where is the $x^k  y^{n-k}$ terms?
You want to sum $\binom{n}{k} $ for all values ${k} \to\ {n}$ the binomial equation can be written 
$\displaystyle\sum\limits_{k=0}^n \binom{n}{k} (x)^k (y)^{n-k} = (x+y)^n$ If you think about your question in terms of finding the sum 
$\displaystyle\sum\limits_{k=0}^n \binom{n}{k}$   ~(The number of possible subsets)
you are actually just multiplying the sum by 1, in which case,  $(x)^k (y)^{n-k}$ must always evaluate to 1 for each of iteration of the summation. this is true when the values of ${x},{y}$ equal 1 so you get $(1+1)^n=2^n$
