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 3 of 9.
Chandrashekar said:
9 years ago
Here, the answer is 20.
By using the fundamental principle of counting ie, if one activity can be done in 'm' num of different ways and the second activity can be done in 'n' number of different ways and these two activities can be done in MxN different ways.
For example, how mane 3 digit number can be formed By using the numbers 2, 3, 5, 8.without repeating.
Soln:-
Let the numbers are 2, 3, 5, 8 and u all know 3 digit number means it must have unit tens and hundred places.
So, now you just Make three column like H T U.
2 3 4 ie in unit place you can fill any of the 4 numbers means 4 different ways and remaining 2 place that is tens and hundred and in the tens place, you can fill in 3 different ways because repetition not allowed.
If once you used the digit you cannot use again. And next 100 places you can fill in 2 different ways.
Therefore total no of ways = 2 x 3 x 4.
= 24ways.
I Hope you understood.
By using the fundamental principle of counting ie, if one activity can be done in 'm' num of different ways and the second activity can be done in 'n' number of different ways and these two activities can be done in MxN different ways.
For example, how mane 3 digit number can be formed By using the numbers 2, 3, 5, 8.without repeating.
Soln:-
Let the numbers are 2, 3, 5, 8 and u all know 3 digit number means it must have unit tens and hundred places.
So, now you just Make three column like H T U.
2 3 4 ie in unit place you can fill any of the 4 numbers means 4 different ways and remaining 2 place that is tens and hundred and in the tens place, you can fill in 3 different ways because repetition not allowed.
If once you used the digit you cannot use again. And next 100 places you can fill in 2 different ways.
Therefore total no of ways = 2 x 3 x 4.
= 24ways.
I Hope you understood.
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.
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.
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.
Peter said:
9 years ago
Thanks a lot for the explanation. It is easy now.
Kalai said:
9 years ago
In the third explanation I didn't understand please explain it.
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.
Pranav Khurud said:
8 years ago
If question is like, How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7, 0, and 9, which are divisible by 5 and none of the digits is repeated?, In this situation unit place may be 0 or 5.
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.
Shivam said:
9 years ago
Why can't we write 5c2? There are two places and 5 numbers.
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