TITLE: Pairs of non-empty disjoint sets
QUESTION [6 upvotes]: Question: 
Let $S=\{1,2, \cdots, 10 \}$. Then the number of pairs $(A, B)$, where $A$ and
$B$ are non-empty disjoint subsets of $S$ is? 
[I could solve the question as demonstrated below, but it involves calculating a tedious sum of products, which would take up a lot of time. It is typically expected to solve this in a couple of minutes, so I was wondering if there was a faster way to do this.]
My approach:
Let the set $A$ consist of $x$ elements. There are ${10 \choose x}$ ways of making that selection. 
We are now left with $10-x$ elements. 
Let the set $B$ consist of $y$ elements. This selection can be done by ${10-x \choose y}$ ways. 
The total number of ways can be found out by summing over the product of the two above as $\sum_{x=1}^{9} \sum_{y=1}^{10-x} {10 \choose x}{10-x \choose y} $ which comes out to be $57002$

REPLY [5 votes]: It is also interesting to note that the sum you get can be simplified in the following way using the binomial theorem:
\begin{align}
\sum_{x=1}^{9}\sum_{y=1}^{10-x}{ 10 \choose x}{ 10-x \choose y} &= \sum_{x=1}^{9}{ 10 \choose x}\left(\sum_{y=0}^{10-x}{ 10-x \choose y} -1 \right)\\
&=\sum_{x=1}^{9}{ 10 \choose x} (2^{10-x} -1) \\
&= \sum_{x=0}^{10}{ 10 \choose x}2^{10-x} -2^{10}-1 - \sum_{x=0}^{10}{ 10 \choose x}+1+1 \\
&= 3^{10}-2^{10}-1-2^{10}+1+1 \\
&= 3^{10}-2^{11}+1
\end{align}
Thus getting a pretty interesting identity.