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 6 of 9.
Chandrashekar pa said:
9 years ago
Divisibility rule 3:
If the sum of the digit is multiple of 3 then that number is divisible by 3.
Ex: 123 is divisible by 3. 1 + 2 + 3 = 6 here 6 is multiple of 3 therefore 123 is divisible by 3.
If the sum of the digit is multiple of 3 then that number is divisible by 3.
Ex: 123 is divisible by 3. 1 + 2 + 3 = 6 here 6 is multiple of 3 therefore 123 is divisible by 3.
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.
Kalai said:
9 years ago
In the third explanation I didn't understand please explain it.
Peter said:
9 years ago
Thanks a lot for the explanation. It is easy now.
Venkat said:
9 years ago
Can any one help me on this one? How many 5 digit numbers can be formed from the digits 0 to 9, so that odd digits are occupied only in even position using one digit only once.
Manpreet said:
9 years ago
How 4 digits are remaining?
(1)
Otieno Bonface Omondi said:
9 years ago
I am not understanding, what has been done. Can someone elaborate the whole thing in a simpler mathematical language? Please.
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.
Ashutosh said:
9 years ago
How many six digit nos can be formed using 3 odd and 3 even numbers if each digit is to be used at most once.
Solve this and find the solution.
Solve this and find the solution.
Justin said:
9 years ago
The best way is,
6c3 = 20.
6c3 = 20.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers