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The $2^n$  vertices of a graph $G$ corresponds to all subsets of a set of size $n$, for $n \geq 6$. Two vertices of $G$ are adjacent if and only if the corresponding sets intersect in exactly two elements.

The maximum degree of a vertex in $G$ is:

  1. $\binom{\frac{n}{2}}{2}.2^{\frac{n}{2}}$
  2. $2^{n-2}$
  3. $2^{n-3}\times 3$
  4. $2^{n-1}$
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If you generalise this question,

I have devised a formula

For intersection to be exactly k elements k<<n

 

Formula is

$(k+1)*2^{n-(k+1)}$

 

In our case it is (k=2)

$3*2^{n-3}$

 

For k=3 it is

$4\times 2^{n-4}$ and so on

You can verify this.
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Let us pick one random vertex of the graph and assume it is a subset S with size k.

According to the question another vertices would be “adjacent if and only if the corresponding sets intersect in exactly two elements”.

Degree of this vertex = Other subsets that has exactly 2 elements in common with S.

For number of adjacent vertices, first we pick 2 elements that are common in both = kC2

Now we cannot take any other element from the rest of the k-2 elements of S1 because then there will be three or more elements common, but we want exactly two.

There are n-k elements remaining, for each of them we can either leave it or take it,

degree(S, k) = (kC2) * 2⁽ⁿ ⁻ ᵏ⁾

Now we have to find the value of k for which  degree(S, k) will be maximum,

The maximum value of degree(S,k) = 3 * 2⁽ⁿ ⁻ ³⁾

[This value appears when k=3 and k=4, and it depends only on k as we can take  2ⁿ common from each term of degree(S,k) ]
Answer:

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