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 9 of 9.
Alok said:
6 years ago
Thanks @Sujit.
(2)
Raghul said:
6 years ago
2 at hundred places there is 4 possibilities.
So, there is 5 numbers (3, 6, 9, 7&2) other than 5, each of them having 4 possibilities. Thus 4*5 - 20.
So, there is 5 numbers (3, 6, 9, 7&2) other than 5, each of them having 4 possibilities. Thus 4*5 - 20.
Dinesh s said:
6 years ago
H T O
_ _ 5.
So two chance only.
Therefore ans 5C2 = 5 x 4 = 20 simple.
_ _ 5.
So two chance only.
Therefore ans 5C2 = 5 x 4 = 20 simple.
(8)
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)
Akash said:
5 years ago
It is 5C2 * 2! = 20.
(8)
Suraj said:
3 years ago
_ _ 5.
There is 2 chance only,
So we use the permutation formula for it,
NPr=n!/ (n-r) !
5P2 = 5!/ (5-2) !
= 5 * 4 * 3 * 2 * 1 /3 * 2 * 1.
= 5 * 4.
= 20.
There is 2 chance only,
So we use the permutation formula for it,
NPr=n!/ (n-r) !
5P2 = 5!/ (5-2) !
= 5 * 4 * 3 * 2 * 1 /3 * 2 * 1.
= 5 * 4.
= 20.
(53)
Moksha said:
2 years ago
We have to form 3-digit numbers which are divisible by 5,
In that case in one's place, there should be 5, there is only one way to place 5 at one's place
So, we have already selected 5 among 2,3,5,6,7,9 then we have only 2,3,6,7,9
there are 5 ways to place a number among 2,3,6,7,9 numbers at ten's place.
So we have to place any number among 2,3,6,7,9 at ten's place, then the remaining numbers we have only 4,
So we have four ways to place any of the numbers in hundred's place.
Therefore the total ways 1*5*4=20.
In that case in one's place, there should be 5, there is only one way to place 5 at one's place
So, we have already selected 5 among 2,3,5,6,7,9 then we have only 2,3,6,7,9
there are 5 ways to place a number among 2,3,6,7,9 numbers at ten's place.
So we have to place any number among 2,3,6,7,9 at ten's place, then the remaining numbers we have only 4,
So we have four ways to place any of the numbers in hundred's place.
Therefore the total ways 1*5*4=20.
(6)
Thanusha K said:
2 years ago
@ Vinay.
If you observe the digits given in question there is no '0'. So at units place, we won't consider '0'. And once that is 5. And each of them has 4 ways.
So, 5 * 4 = 20 ways.
If you observe the digits given in question there is no '0'. So at units place, we won't consider '0'. And once that is 5. And each of them has 4 ways.
So, 5 * 4 = 20 ways.
(1)
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