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.
DharanG said:
9 years ago
There are 6 digits are given ,
As per question one(5) is fixed so 5! = 120.
Three digit number is going to be formed by 3! combination.
So, 120/6 = 20.
As per question one(5) is fixed so 5! = 120.
Three digit number is going to be formed by 3! combination.
So, 120/6 = 20.
Hari said:
2 decades ago
Hi Nikitha,
Here we are considering only 4 digits but not 5 in hundreds place because we already placed a digit in tens place so remaining are only 4.
Here we are considering only 4 digits but not 5 in hundreds place because we already placed a digit in tens place so remaining are only 4.
Java eucy said:
10 years ago
How many even four-digit numbers be formed by using the integers 2, 3, 4, 5 without repetition? How many of these numbers will be less than 3000?
Hanipoo23 said:
1 decade ago
Hey my doubt is not related to this qus. But a general one. What is the diff between permutation and combination in terms of logic, not formulae.
Xander said:
1 decade ago
Here you can fix 3 at the hundred place and 5 at the unit place so only one place is left and 4 digit. So it should be permutation 4 by 1 ans=4.
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.
Ankit Sharma said:
1 decade ago
Friends,
How many 4 lettered words divisible by 4 can be formed fro 0 1 2 3 4 5?
And if repetition of digits is not allowed?
How many 4 lettered words divisible by 4 can be formed fro 0 1 2 3 4 5?
And if repetition of digits is not allowed?
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.
Pranav said:
1 decade ago
How many four digit numbers divisible by 4 can be formed using the digits 2, 3, 6&7 the digits not being repeated?
Xiya said:
8 years ago
Can anyone say is 20 the right answer?
Because the question asked is number divisible by 5 and also not repeated?
Because the question asked is number divisible by 5 and also not repeated?
(1)
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