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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?
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?
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Solution
The correct option is B 30
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 thousands place.The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are: - 0,1and3 as 2+1+0+3=6 is divisible by 3.
- 0,3and4 as 2+3+0+4=9 is divisible by 3.
In both the above combination, the remaining three digits can be arranged in 3! ways. Total number = 2×3!=12
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: - 0,1and2 as 3+1+0+2=6 is divisible by 3.
- 0,2and4 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 = 2×3!=12
Case 3: If we pick 4 for thousands place.The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are: - 0,2and3 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.
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 thousands place.
The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are:
- 0,1and3 as 2+1+0+3=6 is divisible by 3.
- 0,3and4 as 2+3+0+4=9 is divisible by 3.
In both the above combination, the remaining three digits can be arranged in 3! ways. Total number = 2×3!=12
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:
- 0,1and2 as 3+1+0+2=6 is divisible by 3.
- 0,2and4 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 = 2×3!=12
Case 3: If we pick 4 for thousands place.
The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are:
- 0,2and3 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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