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 8 of 9.
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)
Raoakc said:
8 years ago
Can't we do it by combination?
Brown said:
7 years ago
What if the series includes 0, ie 2 3 5 6 7 9 0?
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)
Pooja said:
7 years ago
Well explained. Thanks.
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.
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)
Sarvani said:
6 years ago
Please explain it clearly. I didn't understand.
(2)
Aditya Dubey said:
6 years ago
@Sarvani.
So we have 3 digits, so if the 1s place is 5 we can say the number can be divisible by 5.
Now the number of ways we can fill the 1s place is ( 1 ) _ _ 1W . now we have 2,3,6,7 and 9 so we can fill 100th place in 5 ways.
So we have 5W _ 1W. Now we are left with 10th place and we have only 4 distinct numbers so we can write it as 5W 4W 1W which is nothing but 20.
So we have 3 digits, so if the 1s place is 5 we can say the number can be divisible by 5.
Now the number of ways we can fill the 1s place is ( 1 ) _ _ 1W . now we have 2,3,6,7 and 9 so we can fill 100th place in 5 ways.
So we have 5W _ 1W. Now we are left with 10th place and we have only 4 distinct numbers so we can write it as 5W 4W 1W which is nothing but 20.
(3)
Vinay said:
6 years ago
0 may also come in unit place, please explain me.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers