Answer: 45
AF, EG, DH these three edges will always be part of MST. Now among AE and FG, you can take anyone but not both because then it will create a cycle. Similarly, between ED and GH you can take anyone but not both because then it will create a cycle. In the left-most cycle which is ABCDEA , there are 5 edges you can take any 2 out of them that are not already taken, similarly, for the rightmost cycle, FGHJIF, take any 2 out of them that are not already taken, so till now a total of 3+3=6 edges each edge weight is 7 and 3 edges of weight 1, So total edge weight = 6*7+3*1=45.one of the possible minimum spanning trees will be spanning tree containing edges AF, EG, DH, AE, GH, FI, IJ, AB, BC.
Minimum spanning tree - Wikipedia
The number of possible MST: Between AE and FG you can take anyone so 2 choices, similarly ED and GF have 2 choices. So till now 2*2= 4 choices (AE and ED or AE and GF or FG and ED or FG and GF) between AB, BC, CD you can take any two so 3 choices, similarly between FI, IJ, GH you can take any two, so total 2*2*3*3=36 choices.
