How many unordered pairs $A$ and $B$ are there? Let $S=\{1,2,3,...,10\}$ . How many unordered pairs $A$ and $B$ are there, where $A$ and $B$ are non empty disjoint subsets of $S$ ?
Any approach for such questions ?

Total number of unordered pairs of disjoint subsets of S
This is one related question, but using the same approach as given in the above link, I am getting 
$\left(\frac{3^n - 1}{2} + 1 \right) - 2^n$ for my question

But, I need a confirmation here ?
 A: First assume that we need ordered pairs $A$ and $B$, where $A$ and $B$ are disjoint subsets of $S$. For this, an element can either go to A or B or to neither of them. There are $3^n$ ways of doing this.  
Now let's remove the case when A or B is empty.
Case 1: One is empty, other is not. There are $2 \times (2^n - 1)$ such cases.
Case 2: Both are empty: There is only $1$ such case.
We need to remove these $ 2^{n+1} - 1 $ cases from $3^n$.
So we are left with $3^n - 2^{n+1} + 1$ possibilities.
But since we need unordered pairs of $A$ and $B$, we divide this by $2$ to get $\boxed{\frac{3^n + 1}{2} - 2^n}$.
One approach for solving such questions is to start with a broader case that has less restrictions and then apply restrictions on it.  
A: Here is a proof of your second question.
Consider a set $E$ containing $n$ elements ($\mathrm{card}(E)=n$) and $q=\mathrm{card}(F)$ where :
$$F=\{(A,B)\in\mathcal{P}(E)^2;\,A\cap B=\emptyset\}$$
Since the single subset $A$ of $E$ verifying $A\cap A=\emptyset$ is $\emptyset$ itself, the answer to your question will be $1+\frac{q-1}{2}$.
Since the elements of $F$ can be classified according to their cardinal, we have :
$$F=\bigcup_{k=0}^nF_k\qquad\mathrm{where}\quad F_k=\{(A,B)\in\mathcal{P}_k(E)^2;\,A\cap B=\emptyset\}$$
where $\mathcal{P}_k(E)$ denotes the set of all subsets of $E$ whose cardinal is $k$.
This is the union of pairwise disjoint sets, and therefore :
$$q=\sum_{k=0}^n\mathrm{card}(F_k)$$
We see that $\mathrm{card}(F_k)={n\choose k}2^{n-k}$. Thus :
$$q=\sum_{k=0}^n{n\choose k}2^{n-k}=3^n$$
Finally, there are $1+\frac{3^n-1}{2}$ unordered pairs of disjoint subsets.
A: We reformulate the problem somewhat and verify OPs formula
\begin{align*}
\frac{3^n - 1}{2} + 1 - 2^n
\end{align*}
by  showing that the  question asks  in  fact for   famous numbers.

We consider  the set $S_n=\{1,2,3,\ldots,n\}$ and ask for the number of non-empty partitions of the set 
  \begin{align*}
S_{n+1}=S_n\cup\{n+1\}
\end{align*}
  into three non-empty 
  sets.

Note, this number is the same as OPs question  is  asking  for, because whenever we consider an unordered pair $(A,B)$ with $A,B\subset S_n,  A\cap B=\emptyset$ we have one of two cases:


*

*$A\cup B=S_n$: This corresponds to  the partition $S_{n+1}=A\cup B\cup\{n+1\}$ and vice versa.

*$A\cup B \ne S_n$: This corresponds to the partition $S_{n+1}=A\cup B\cup \left(S_{n+1}\setminus (A\cup B)\right)$ and vice versa.
The number of ways of partitioning a set of $n$ elements into $k$ nonempty sets is called Stirling number of the second kind and denoted with ${n\brace k}$. They admit the  representation
\begin{align*}
{n\brace k}=\frac{1}{k!}\sum_{j=0}^k(-1)^{k-j}\binom{k}{j}j^n
\end{align*}

Here we are looking for ${n+1\brace 3}$, the number of ways of partitioning a set of $n+1$ elements into $3$ nonempty sets. We obtain for $n>0$:
  \begin{align*}
{n+1\brace 3}&=\frac{1}{3!}\sum_{j=1}^3(-1)^{3-j}\binom{3}{j}j^{n+1}\\
&=\frac{1}{6}\left(\binom{3}{1}1^{n+1}-\binom{3}{2}2^{n+1}+\binom{3}{3}3^{n+1}\right)\\
&=\frac{1}{2}\left(3^{n}+1\right)-2^n
\end{align*}
  which coincides with OPs result.

Note: The sequence $$\left({n\brace 3}\right)_{n\geq 0}=(0,0,0,1, 6, 25, 90, 301, 966, 3025, 9330, 28501,\ldots)$$ is archived in OEIS as A000392.
