TITLE: Total number of unordered pairs of disjoint subsets of S
QUESTION [25 upvotes]: Let $S = \{1, 2, 3, 4\}$. Find the total number of unordered pairs of
  disjoint subsets of $S$.

I know the answer is $41$ since it's solved in the book as the expression 
$$\frac{3^4 -1}{2!} +1 \ .$$ 
I couldn't get through the solution above, I have solved it by making pair of sets. Please help me to understand the way by which the question has been solved in the book.

REPLY [9 votes]: Assuming, $\large\color{maroon}{\text{S}}$ = $\left \{ 1,2,3 \right \}$ , then following is the required Relation.
$$\begin{align*} \text{Z} = \left \{ \left ( R,T \right ) \;\; | \; \; \text{R,T} \subseteq \; \text{S ,} R \cap T = \phi \right \} \end{align*}$$
The pair $(R,T)$ is a disjoint subset pair of $S$.
$Z$ in matrix representation, $\\$

First we will count the cardinality of $Z$ based on ordered pair $(R,T)$.
example :

first green circle means $\left \{ 2 \right \}$ and $\left \{ 3 \right \}$ is one pair $\left ( \left \{ 2 \right \},\left \{3  \right \} \right )$ in the relation $Z$. And because, we are considering ordered pairs ,then $\left ( \left \{ 3 \right \},\left \{2  \right \} \right )$ will also be a pair in relation $Z$.
Second green circle means $\left \{ 2,3 \right \}$ and $\left \{ 1 \right \}$ is one pair $\left ( \left \{ 2,3 \right \},\left \{1  \right \} \right )$ in the relation $Z$. And we are considering ordered pairs, then $\left ( \left \{ 1 \right \},\left \{2,3  \right \} \right )$ will also be a pair in relation $Z$.

Now,
No of such ordered pairs in relation $Z$ =  No of $1$'s in the matrix representation.
Counting procedure:
one example: 

2nd row of matrix, which is related to set $\left \{ 1 \right \}$.
No of subsets that $\left \{ 1 \right \}$ will be related to are the subsets of $\left \{ 2,3 \right \}$.
How many subsets of $\left \{ 2,3 \right \}$ ? . = $2^{3-1}$ .(because $1$ is missing here).
Similarly for subsets $\left \{ 2 \right \}$ and $\left \{ 3 \right \}$ ,, i.e. 3rd and 4th row of the matrix.
So total $1$'s for these $\binom{3}{1} = 3$ ,  one element subsets $\left ( \left \{ 1 \right \},\left \{ 2 \right \},\left \{ 3 \right \} \right )$ , is = $\binom{3}{1}*2^{3-1}$. (Or, total no of 1's in 2nd,3rd,4th rows of the matrix)

Doing in the same way, will arrive in the follwoing series,
$$\begin{align*} &=\binom{3}{0}*2^{3-0} + \binom{3}{1}*2^{3-1} + \binom{3}{2}*2^{3-2} + \binom{3}{3}*2^{3-3} \\ &=\binom{3}{0}*1^0*2^{3-0} + \binom{3}{1}*1^1*2^{3-1} + \binom{3}{2}*1^2*2^{3-2} + \binom{3}{3}*1^3*2^{3-3} \\ &=\sum_{r=0}^{3}\binom{3}{r}1^{r}.2^{3-r} \\ &= (1+2)^{3} \\ &=3^3 \end{align*}$$ 
If we assume $\large\color{maroon}{\text{S}}$ = $\left \{ 1,2,3,4,5,6....n \right \}$
Then this above value will be $\large\color{red}{3^n}$
Now, how many unordered ?

This relation Z is symmetric with only one self-loop at $\phi$..
Whenever we need only unordered pairs $(R,T)$ then, we will only count off-diagonal $1$'s and one $(\phi,\phi)$ pair.

No of unordered $(R,T)$ pairs = 
 $\begin{align*} \frac{3^n -1}{2} +1 \end{align*}$