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....