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:
89 comments Page 1 of 9.
Amrutha said:
4 weeks ago
Someone asked that 0 may also come in the units place.
But in the given numbers, we have only 2, 3, 5, 6, 7, 9.
From these numbers only we want form 3-digit numbers; 0 is not given.
So, no need to take 0.
But in the given numbers, we have only 2, 3, 5, 6, 7, 9.
From these numbers only we want form 3-digit numbers; 0 is not given.
So, no need to take 0.
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)
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)
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.
(54)
Akash said:
6 years ago
It is 5C2 * 2! = 20.
(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)
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)
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.
Alok said:
6 years ago
Thanks @Sujit.
(2)
Vinay said:
6 years ago
0 may also come in unit place, please explain me.
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