The total number of three-digit numbers divisible by 3, which can be formed using the digit 1, 3, 5, 8 if repetition of digit is allowed, is

All are different formed by (1, 3, 5) is 3!, (1, 3, 8) is 3!
sum = 12 
Two repetitions formed by (5, 5, 8)is $\frac{3!}{2!}$, (8, 8, 5) is $\frac{3!}{2!}$ 
Sum =6 
Three repetitions = 4 (111, 333, 555, 888)
Total number of numbers = 12 + 6 + 4 = 22