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 5 of 9.
Brown said:
7 years ago
What if the series includes 0, ie 2 3 5 6 7 9 0?
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.
Vinay said:
6 years ago
0 may also come in unit place, please explain me.
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.
Deepu said:
1 decade ago
Karthikas answer is the most easy to understand and generic concept. I recommend to follow that method.
Sujit said:
1 decade ago
@Suganya.
In number 456.
6 is in Unit Place.
5 is in 10th place.
4 is in 100th place.
Hope you get this :-).
In number 456.
6 is in Unit Place.
5 is in 10th place.
4 is in 100th place.
Hope you get this :-).
Suganya said:
1 decade ago
Please explain the 1s 10s &100 place in number detaily.
Ayodele k said:
1 decade ago
If the three digits must be divisible by 5 and no digit is to be repeated, therefore, it's safe to reserve a spot for 5 and then have 2,3,6,7,9 combine the first two digits (XY5),
Permutation of the first two digits gives 5C2
5!/3!
23,26,27,29
32,36,37,39
62,63,67,69
72,73,76,79
92,93,96,97
These outcomes combine with 5 each. 5P3*1 resulting in 20 outcomes.
Remember, no repetition, so there
Permutation of the first two digits gives 5C2
5!/3!
23,26,27,29
32,36,37,39
62,63,67,69
72,73,76,79
92,93,96,97
These outcomes combine with 5 each. 5P3*1 resulting in 20 outcomes.
Remember, no repetition, so there
Chandra said:
1 decade ago
n!/ (n-r) ! is the best formula and easy way to get answer.
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