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 4 of 9.
Karthick said:
9 years ago
How many 4 digit no can be formed with digits 1, 2, 3,4,5 which are divisible by 4 and digits not repeated?
a)144 b) 168 c) 182 d) none
Please tell me the correct answer.
a)144 b) 168 c) 182 d) none
Please tell me the correct answer.
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.
Vijay015125 said:
9 years ago
On units place 5 is fixed.
On 10's place remaining 5 digits one is selected = 5.
At finally 100's place remaining 4digits will be there so 4.
RESULT = 4 * 5 * 1 = 20.
On 10's place remaining 5 digits one is selected = 5.
At finally 100's place remaining 4digits will be there so 4.
RESULT = 4 * 5 * 1 = 20.
Sheraz said:
1 decade ago
How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated in a number?
Siva said:
1 decade ago
Lets see the detail combination divisible by 5 are 235, 265, 275, 295.
Similarly for other numbers. Hence there are 5 number and 4 combination for each number.
Similarly for other numbers. Hence there are 5 number and 4 combination for each number.
Shantanu said:
8 years ago
How many five digit numbers can be formed by using digits 0, 1, 2, 3, 4 and 5 without repetition. And number should be divisible by 6.
Can anyone solve this?
Can anyone solve this?
Ayush jha said:
9 years ago
How many 5-digited numbers can be formed by using 0, 1, 2, 3, 4, 5, 6 which is divisible by 7 when repetition is allowed?
Please help me to get the answer.
Please help me to get the answer.
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)
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)
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.
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