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 4 of 9.
Manisha said:
9 years ago
Nice answer@Justin.
Monisha said:
9 years ago
@Karthick.
Since last two digits must be 12 or 24 or 32 or 52,
The other two digits must be filled with either of the 3 numbers,
Thus,
(3*2)+(3*2)+(3*2)+(3*2) = 24 ways.
And hence the answer is none.
Since last two digits must be 12 or 24 or 32 or 52,
The other two digits must be filled with either of the 3 numbers,
Thus,
(3*2)+(3*2)+(3*2)+(3*2) = 24 ways.
And hence the answer is none.
Joe said:
9 years ago
Why isn't it just 5C2, from the 5 numbers that arent 5 to go in tens and hundreds, and picking two?
Vignesh said:
9 years ago
Please explain me, what would happen if another digit (0) is added to the series, i.e., 2,3,6,7,9,5 and 0 and the digits are to arranged without repetition such that the 3 digit number is divisible by 5?
Ambika said:
8 years ago
Write all the possible number using the digit 7, 0, 6 repetition of digits is not allowed.
Dema said:
8 years ago
How 5p2? Please explain.
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.
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?
Raoakc said:
8 years ago
Can't we do it by combination?
Brown said:
7 years ago
What if the series includes 0, ie 2 3 5 6 7 9 0?
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