$S = {1,2,3,4,5,6.....n}$

Let us assume any two subset $S_1$ and $S_2$. We can simply assume $n(S_1 \cap S_2) =0$ to consider the disconnected sets if we want.

Now there are three cases in which $(S_1 \backslash S_2) \cup (S_2 \backslash S_1) \;\; Or, \;\; (S_1 \oplus S_2) $ has only $2$ element.

- Both green shaded area has one element each and in this case sizes of $S_1$ and $S_2$ are same.
- The green area of $S_1$ contains $2$ element and the green area of $S_2$ contains none. In this case size of $S_1$ is $2$ more than that of $S_2$.
- The green area of $S_2$ contains $2$ element and the green area of $S_1$ contains none. In this case size of $S_2$ is $2$ more than that of $S_1$.

So, if we are only interested in a **particular **set vertex corresponding to set $S_1$ of size $= m$, then $S_1$ is connected to **three types** of set vertices as shown below. We will use the words "**set**" and "**vertices**" synonymously.

In this above image, we have considered $m \geq 2$. The cases for $m = 1 \text{ and } m = 0$ will be **discussed later**.

Now, what we need to find is the **no of set vertices** in each of the **above** three types and sum them up to get the **degree** of the vertex corresponding to the set $S_1$.

For simplicity let us assume $S = \{1,2,3,4,5,6,7\}$ and set $S_1 = \{1,2,3,4\}$. Our interest will be to find $S_2$ such that vertices corresponding to $S_1$ and $S_2$ are connected.

**CASE 1** : If we try to find another set $S_2$ having $4$ elements and satisfying constraint $n(S_1 \oplus S_2) = 2$, then we will see that no of such set $S_2$ is $4 \cdot (7 - 4)$. Or in general if $S_1$ is an $m$ element set then no of such $S_2$ sets with constraint $n(S_1 \oplus S_2) = 2$ will be equal to $m\cdot (n-m)$.
**CASE 2 **: $S_1$ contains $4$ element and If we try to find $S_2$ where $S_2$ contains $2$ elements and satisfying constraint $n(S_1 \oplus S_2) = 2$, then no of such $S_2$ will be $4C2$ or in general, for $m$ element set $S_1$, we have $mC2$ no of $S_2$ type sets all with $(m-2)$ size.
**CASE 3**: $S_1$ contains $4$ element and If we try to find $S_2$ where $S_2$ contains $6$ element and satisfying constraint $n(S_1 \oplus S_2) = 2$, then no of such $S_2$ sets will be $3C2$ or $(7-4)C2$. In general, with $S_1$ being $m$ element set, then $(n-m)C2$ no of $S_2$ sets will be possible.

Therefore, summing all three cases :

Degree of vertex $S_1$ ( assuming general case of $n(S_1) = m$ )

$\begin{align*}

&=m\cdot (n-m) + \binom{m}{2} + \binom{n-m}{2} \\

&=m\cdot n - m^2 + \frac{m^2}{2} - \frac{m}{2} + \frac{(n-m)\cdot (n-m-1)}{2} \\

&=m\cdot n - m^2 + \frac{m^2}{2} - \frac{m}{2} + \frac{n\cdot (n-1)}{2} \\

&\qquad - \frac{n \cdot m}{2} - \frac{n \cdot m}{2} + \frac{m^2}{2} + \frac{m}{2} \\

&=\frac{n\cdot (n-1)}{2} \\

&=\binom{n}{2} \\

\end{align*}$

This result is independent of $m$ for $m \geq 2$ and $m \leq n$.

For $m = 0$ and $m = 1$ also we can show that degree of $0$ and $1$ size set vertices is nothing but $nC2$ only. (fairly straight forward cases).

So we can conclude that every vertex has the same degree and the degree is $nC2$.

Now we can guess one thing by looking at the following image:

i.e.for $m \geq 2$ if $m$ is even the $S_1$ is connected to only even cardinality type of sets (**at least one**) or if $m$ is odd then $S_1$ is connected to only odd cardinality type of sets (**at least one**). By this, we can almost say that there are two connected components in the graph.

But there is little more argument before we can proceed and have a valid proof.

if $m = 0$ then $S_1 = \phi$, Then $S_1$ will be connected to all $m = 2$ type of sets or $2$ cardinality sets.

if $m = 1$ then $S_1$ will be one of all $1$ element sets, Then $S_1$ will be connected to all other $1$ cardinality sets and at least one $3$ cardinality set.

We can argue that, one $m$ (even) cardinality set is at least connected to one $(m-2)$ cardinality set. That particular $(m-2)$ cardinality set is at least connected to one $(m-4)$ cardinality set and so on till $\phi$ set vertex. There for all even cardinality sets are connected to $\phi$ directly or indirectly.

A similar argument holds for odd cardinality set vertices till we reach some $1$ cardinality set. Moreover all $1$ cardinality sets are connected.

Therefore we have a situation now that all even cardinality sets form one connected component and all odd cardinality set form another component.

For example : $n = 4$ :