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?

पर्याय

• 30

• 48

• 24

• 36

Answer 1

30

Explanation:

The thousands place can only be filled with 2, 3 or 4, since the number is greater than 2000.

For the remaining 3 places, we have pick out digits such that the resultant number is divisible by 3.

It the sum of digits of the number is divisible by 3, then the number itself is divisible by 3.

Case I: If we take 2 at thousands place.

The remaining digits can be filled as:

0, 1 and 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 combinations the remaining three digits can be arranged in 3! ways.

∴ Total number of numbers in this case = 2 × 3! = 12.

Case II: If we take 3 at thousands place. The remaining digits can be filled as:

0, 1 and 2 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 combinations, the remaining three digits can be arranged in 3! ways. Total number of numbers in this case = 2 × 3! = 12.

Case III: If we take 4 at thousands of place.

The remaining digits can be filled as:

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 of numbers in this case = 3! = 6.

∴ Total number of numbers between 2000 and 5000 divisible by 3 is 12 + 12 + 6 = 30.