The number of numbers divisible by 3 that can be formed by four different even digits is
(a) 18
(b) 36
(c) 0
(d) None of these

Hint: For finding the number of numbers divisible by 3 we use the permutation concept. First, we must know the divisibility rule of 3 which states that a number is divisible by 3 only when the sum of individual digits is divisible by 3. By using this we can easily solve our problem.

Complete step-by-step answer:

According to our problem, even digits can be stated as 0, 2, 4, 6, 8.

As the divisibility rule of 3 states that a number is divisible by 3 only when the sum of individual digits is divisible by 3. So, the 4 different even digit numbers are divisible by 3 = set of (2, 4, 6, 0) and (8, 6, 4, 0).
So, the number of combinations possible = $4!$.
But, if 0 occurs at first place then the number is not four digits. So, cases for 0 at first place = $3!$.
So, total valid combinations = $4!-3!=18$.
But there are two possible sets, so multiplying the obtained result by 2 we get total valid combinations as = $18\times 2=36$.
Therefore, option (b) is correct.

Note: The key concept of solving this problem is the knowledge of permutations and divisibility rule for 3. Once the total number of cases are obtained by using the given criteria, then by using permutation the final result can be evaluated without any error.