A) 30
B) 48
C) 24
D) 36
Correct Answer: A
Solution :
There are 4 places to be filled with the given digits. The thousands place can have only 2, 3 and 4 since the number has to be greater than 2000. For the remaining 3 places, we have pick out digits such that the resultant number is divisible by 3. The divisibility criteria for 3 states that sum of digits of the number should be divisible by 3. Case 1: if we pick 2 for thousand place. The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are: \[0,1\]ad 3 as 2 + 1 + 0 + 3 = 6 is divisible by 3. 0,3 and 4 as 2 + 3 + 0 + 4 = 9 is divisible by 3. In both the above combination, the remaining three digits can be arranged in ways. Total number Case 2: If we pick 3 for thousands place The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are: 0,1and as 3 + 1 + 0 + 2 = 6 is divisible by 3. 0,2 and 4 as 3 + 2 + 0 + 4 = 9 is divisible by 3. In both the above combination, the remaining three digits can be arranged in 3! ways. Total number \[=2\times 3!=12\] Case 3: if we pick 4 for 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 as 4 + 2 + 0 + 3 = 9 is divisible by 3. In the above combination, the remaining three digits can be arranged in 3! ways. Total number = 3! = 6 Total number of numbers between 2000 and 5000 divisible by 3 are 12 + 12 + 6 = 30.Option A is correct.
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