Question

The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is ?
A. 30
B. 48
C. 24
D. 36

Answer 1

Hint: Let us find those numbers whose sum of all digits is divisible by 3 because if the sum of all digits of a number is divisible by 3 then the number will also be divisible by 3.

Complete step-by-step answer:
As we know that there are 4 places to be filled with the given digits. And the thousands place can only have 2, 3 and 4 since the number has to be greater than 2000 and less than 5000. For the remaining 3 places, we had to pick out digits such that the sum of digits is divisible by 3.

So, there will be three possible cases.


Case 1: If we pick digit 2 for the thousand place.

The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are:
0, 1 and 3 because 2 + 1 + 0 + 3 = 6 is divisible by 3.

So, three digits 0, 1 and 3 can be arranged in 3! ways = 6 ways

0, 3 and 4 because 2 + 3 + 0 + 4 = 9 is divisible by 3.

So, three digits 0, 3 and 4 can be arranged in 3! ways = 6 ways

So, total numbers possible with 2 as its thousand place and lying between 2000 and 5000 and divisible by 3 is 6 + 6 = 12.


Case 2: If we pick digit 3 for the thousand place.

The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are:
0, 1 and 2 because 3 + 0 + 1 + 2 = 6 is divisible by 3.

So, three digits 0, 1 and 2 can be arranged in 3! ways = 6 ways

0, 2 and 4 because 3 + 0 + 2 + 4 = 9 is divisible by 3.

So, three digits 0, 2 and 4 can be arranged in 3! ways = 6 ways

So, total numbers possible with 3 as its thousand place and lying between 2000 and 5000 and divisible by 3 is 6 + 6 = 12.


Case 3: If we pick digit 4 for the thousand place.

The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are:
0, 2 and 3 because 4 + 0 + 2 + 3 = 9 is divisible by 3.

So, three digits 0, 2 and 3 can be arranged in 3! ways = 6 ways

So, total numbers possible with 4 as its thousand place and lying between 2000 and 5000 and divisible by 3 is 6

Total number of numbers between 2000 and 5000 divisible by 3 will be the sum of total possible numbers in all three cases and that is 12 + 12 + 6 = 30

Hence, the option A will be the correct option.

Note: Whenever we come up with this type of problem then first, we select the possible digit for the thousand place and then divide it into different cases according to that. Then we had to find total, different ways to place other digits in ones, tens and the hundred places such that the sum of all the digits of the number is divisible by 3.