On observing the graph carefully we will find that for any MST of given graph ,
we have to select 2 edges from first cycle A-J-B as 3C2 , then one edge from AC or BC as 2C1, one from DF and EF as 2C1 , one from FG and FH as 2C1 and then one from cycle G-I-H as 3C2 .
Hence total MST possible =3 X 2 X 2 X 2 X 3= 72