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 7 of 9.
Poonam said:
1 decade ago
Hi friends.
The remaining 5 digits can be arranged in 5! ways or 5p2 ways.
The remaining 5 digits can be arranged in 5! ways or 5p2 ways.
Shree Hemalatha said:
1 decade ago
How the 4 digits are coming in 100's place?
Could you explain me that ?
Could you explain me that ?
Kalai said:
9 years ago
In the third explanation I didn't understand please explain it.
Chandra said:
1 decade ago
n!/ (n-r) ! is the best formula and easy way to get answer.
Suganya said:
1 decade ago
Please explain the 1s 10s &100 place in number detaily.
Shivam said:
9 years ago
Why can't we write 5c2? There are two places and 5 numbers.
Shanthi said:
1 decade ago
Please explain 1s, tens, hundreds place numbers in detail.
Xyz said:
6 years ago
@Dinesh.
It is 5P2 = 5*4 =20.
If it is 5C2 = 5*4/2*1 =10.
It is 5P2 = 5*4 =20.
If it is 5C2 = 5*4/2*1 =10.
(5)
Surbhi srivastava said:
1 decade ago
The greatest 4 digit number which is only divisible by 5.
Joel said:
10 years ago
I need some help on how to come up with those digits?
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