The total number of three-digit numbers, divisible by 3 , which can...

Given,

Three-digit number to be formed which are divisible by $3$ using the digits $1, 3, 5, 8$ and repetition is allowed,

Now taking case \left(1\right)$, where all digits are same, we get</p><p>\left(1, 1, 1\right), \left(3, 3, 3\right), \left(5, 5, 5\right), \left(8, 8, 8\right) \to 4\text{ways}$

Now taking case \left(2\right)$, where$2 -$digit are same and one is distinct, we get</p><p>\left(5, 5, 8\right) \to \dfrac{3 !}{2 !} = 3\text{ways,}\left(8, 8, 5\right) \to 3\text{ways}$

Now taking case \left(3\right)$, where all are distinct, we get</p><p>\left(1, 3, 5\right) \to 6\text{ways}, \left(1, 8, 3\right) \to 6\text{ways}$

So, total ways will be $4 + 3 + 3 + 6 + 6 = 22$