Best way to solve this for GATE is to take $n=2$ and $n=3$ and we get degree of each vertex = ${}^nC_2$ and no. of connected components = $2$.

Lets do it more formally.

It is clear {} should get connected to all $2$ element subsets (and not to any other) of $S$. So, degree of the corresponding vertex is ${}^nC_2$ as we have ${}^nC_2$ ways of choosing 2 elements from $n$. So, answer to first part must be this as it is given degree of all vertices are same.

Now, for the second part, from the definition of $G$ all the vertices of cardinality $k$ will be disconnected from all the vertices of cardinality $k-1$. This is because either all the $k-1$ elements must be same in both or $k-2$ elements must be same in both or else the symmetric difference will be more than $2$. Now if $k-1$ elements are same, symmetric difference will just be $1$. If $k-2$ elements are same, we have one element in one set not in other and $2$ elements in other set not in this, making symmetric difference $3$. Thus symmetric difference won't be $2$ for any vertices of adjacent cardinality making them disconnected.

All the vertices of same cardinality will be connected - when just one element differs. Also, vertices with cardinality difference 2 will be connected- when 2 new elements are in one vertex. Thus we will be getting $2$ connected components in total.

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