Pairs of non-empty disjoint sets

Suppose we have set $A$, $B$ and $S=\{1,2,...10\}$. So for each $x\in S$ you can put it in exactly one of this sets: $A$ or in $B$ or $(A\cup B)'$. So for each $x$ you have 3 choices and thus you can choose $A,B$ on $3^{10}$ ways.

Now we have to substract all pairs where one of the sets is empty. If $A$ is empty, $B$ could be any subset apart from $A$. So we have $2^{10}$ choices. The same is true if $B$ is empty. So we have $2^{11}$ such pair of sets. But we have pair $(\emptyset,\emptyset )$ counted twice, we have to substract $2^{11}-1$ bad pairs of sets.

So the finaly result is $3^{10}-2^{11}+1$

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.