Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 7)
7.
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Answer: Option
Explanation:
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 x 5 x 4) = 20.
Discussion:
88 comments Page 2 of 9.
Sarvani said:
6 years ago
Please explain it clearly. I didn't understand.
(2)
Pritesh said:
6 years ago
1. 5 should come at unit place.
2. Remaining two i.e. tens and hundred places should be filled by remaining five nos. So we need to first select two nos. Among five i.e 5C2.
3. Arrange those selected two nos = 2!.
4. So, the required Answer = 5C2*2!=20.
2. Remaining two i.e. tens and hundred places should be filled by remaining five nos. So we need to first select two nos. Among five i.e 5C2.
3. Arrange those selected two nos = 2!.
4. So, the required Answer = 5C2*2!=20.
(1)
Shaurya said:
7 years ago
To divide with the no 5, no must have 5 at its unit place.
So 5 is fixed at a unit place now there are 5 digits that are left and we have to choose only 2 of them so 5p2.
So 5 is fixed at a unit place now there are 5 digits that are left and we have to choose only 2 of them so 5p2.
Pooja said:
7 years ago
Well explained. Thanks.
Sunthar said:
7 years ago
(23679) 5.
The format should be like XY5.
Since 5 is fixed in Once position so that only outcome number will be divisible by 5.
Out of 3 positions, 2 only left.
So that 2 numbers from 5 numbers (2, 3, 6, 7, 9) can occupy the 2 spots.
(i.e) 5p2 = 5!/(5-2)! = 120/6 = 20.
5 remaining numbers. 2 spots.
The format should be like XY5.
Since 5 is fixed in Once position so that only outcome number will be divisible by 5.
Out of 3 positions, 2 only left.
So that 2 numbers from 5 numbers (2, 3, 6, 7, 9) can occupy the 2 spots.
(i.e) 5p2 = 5!/(5-2)! = 120/6 = 20.
5 remaining numbers. 2 spots.
(1)
Brown said:
7 years ago
What if the series includes 0, ie 2 3 5 6 7 9 0?
Raoakc said:
8 years ago
Can't we do it by combination?
CHALS said:
8 years ago
Can anyone answer for this?
How many can 5 digit number divisible by 4 be formed using the digits 1, 2, 3, 4, 5, 6 no digits being repeated in the numbers?
How many can 5 digit number divisible by 4 be formed using the digits 1, 2, 3, 4, 5, 6 no digits being repeated in the numbers?
(1)
Shantanu said:
8 years ago
How many five digit numbers can be formed by using digits 0, 1, 2, 3, 4 and 5 without repetition. And number should be divisible by 6.
Can anyone solve this?
Can anyone solve this?
Pranav Khurud said:
8 years ago
If question is like, How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7, 0, and 9, which are divisible by 5 and none of the digits is repeated?, In this situation unit place may be 0 or 5.
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