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How many 3 digit number can be formed with the digits 5, 6, 2, 3, 7 and 9 which are divisible by 5 and none of its digit is repeated?
- a)12
- b)16
- c)20
- d)24
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
How many 3 digit number can be formed with the digits 5, 6, 2, 3, 7 an...
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 * 5 * 4) = 20.
Most Upvoted Answer
How many 3 digit number can be formed with the digits 5, 6, 2, 3, 7 an...
Problem: Finding the number of 3 digit numbers that can be formed with the digits 5, 6, 2, 3, 7 and 9 which are divisible by 5 and none of its digit is repeated.
Solution:
To be divisible by 5, the number must end with either 5 or 0. Since none of the digits can be repeated, there are only 2 possible numbers that can be formed with 5 or 0 at the end. So, we need to find the number of ways to form a 3 digit number with the remaining digits such that none of them is repeated.
- Case 1: Numbers ending with 5
If the number ends with 5, then the first two digits can be any of the remaining 4 digits (6, 2, 3, 7). Therefore, the total number of such numbers is 4P2 (permutation of 2 digits from 4) = 12.
- Case 2: Numbers ending with 0
If the number ends with 0, then the first two digits can be any of the remaining 3 digits (6, 2, 3, 7). Therefore, the total number of such numbers is 4P2 (permutation of 2 digits from 4) = 12.
Therefore, the total number of 3 digit numbers that can be formed with the given digits and are divisible by 5 and none of its digits is repeated is 12 + 12 = 24.
Hence, the correct option is (d) 24.
Solution:
To be divisible by 5, the number must end with either 5 or 0. Since none of the digits can be repeated, there are only 2 possible numbers that can be formed with 5 or 0 at the end. So, we need to find the number of ways to form a 3 digit number with the remaining digits such that none of them is repeated.
- Case 1: Numbers ending with 5
If the number ends with 5, then the first two digits can be any of the remaining 4 digits (6, 2, 3, 7). Therefore, the total number of such numbers is 4P2 (permutation of 2 digits from 4) = 12.
- Case 2: Numbers ending with 0
If the number ends with 0, then the first two digits can be any of the remaining 3 digits (6, 2, 3, 7). Therefore, the total number of such numbers is 4P2 (permutation of 2 digits from 4) = 12.
Therefore, the total number of 3 digit numbers that can be formed with the given digits and are divisible by 5 and none of its digits is repeated is 12 + 12 = 24.
Hence, the correct option is (d) 24.
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