Total cases will be 5C4×4!
Now favourable cases for 2f(b)=f(c)+f(d)−f(a) will be,
Case (I) if f(b)=1 then f(c),f(d),f(a) can take the value 3,4,5 & 4,3,5 respectively
Case (II) if f(b)=2 then f(c),f(d),f(a) can take value 3,5,4&5,3,4 respectively
Case (III) if f(b)=3 then f(c),f(d),f(a) can take value 2,5,1&5,2,1 respectively
So total favourable case will be 6
So probability=totaloutcomesfavourablecases=5×4!6=201