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I am getting 128 as answer:

8 Spanning trees in first part(LHS) * 2 choice in middle Part * 8 Spanning trees in Second part(RHS), but given answer is 256:

Actually its calculation is cumbersome..We have to ensure that double counting does not occur..For that proper enumeration of cases is necessary which will be a tedious process for this..

+1 vote
Hi, actually you can select e and f edges together also which will make more number of spanning trees.

128  ( either you select e or f )  +  extra spanning tree ( selecting e and f both ).

One  way to calculate total number of distinct spanning tree in an undirected unweighted graph having no simple loop and no multiple edges and connected using Laplacian matrix.

L(G) =  D - A

Where D is degree matrix  of N * N  and A adjacency matrix of N * N

if( i == j)  then D[i][j] =   total number of edges incident on ith node

else   if(i != j )  then D[i][j] = 0        ( Hence only diagonal elements have some value >= 0 ) rest all are zero

A :  adjacency matrix  ,   A[i][j] =  1  if  node i and node j have edge between them else A[i][j] = 0

Now Laplacian matrix L(G) =  D - A

Now total number of disctint spanning tree in the graph  is   =    (-1)^(i + j ) - cofactor after removing ith row and jth column

where you can take i and j , any values between 1 and N.

means calculating any one co-factor of matrix will give you the number of distinct spanning tree in the graph.

But helpfull only for small graph, because it's difficult for us to calculate determinant of large matrix by hands....
yes you're correct we can select both which will give more number of spanning trees but applying krrichof's theorem as you said is very difficult for such a huge graph :( :(
Then just fix e and f edges,  and select two edges from left side and make left portion connected , and check number of ways  to select from right part.

Then select another way for left part for selecting two edges and multiply it number of ways to select two edges from right part.