States = x, y, z input alphabet = (a, b)
Initial state = x
(x,a) = 3 ( In state x, on input a, transition can be either to x or y or z, 3 possibilities.
(x,b) = 3
(y,a) = 3
(y,b) = 3
(z,a) = 3
(z,b) = 3
total possible DFAs = 3^6 = 729
now we have to choose final state, there are three states and final states is a subset of Q
hence there can be upto 8 final states.
Total DFAs = 729*8 =5832
