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4 votes
Suppose we formed a simple graph with 8 vertices.

Then it will have one unique adjacency matrix representation.

So, the number of simple graphs with 8 vertices = no. of 8×8 symmetric matrices whose elements are either 0 or 1 except the diagonal elements which are zero always.

Now, a 8×8 matrix has 64 cells. Substracting the diagonal cells we have 56 cells. On the upper half of the matrix has (56/2) = 28 cells

Now, for each of the 28 cells we have 2 options 1 or 0.

So,the number of 8×8 symmetric matrices where the elements are either 0 or 1 except the diagonal elements are 2^28

So,the number of simple graphs with 8 vertices is 2^28.
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Number of ways of selecting edges= n(n-1)/2

Now, every edge can be selected or rejected,So for each edge we have 2 ways of selection.

So, for n(n-1)/2 edges , we have 2^(n(n-1)/2) of selecting i.e 2^28

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