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.
Pravas said:
9 years ago
Shortcut way to solve this problem :-6C3 => 6!/3!
= 6 * 5 * 4/3 * 2 * 1.
= 20.
= 6 * 5 * 4/3 * 2 * 1.
= 20.
Akhil said:
9 years ago
Thanks @Rahul! good explanation.
Shivam said:
9 years ago
Say number 5 is placed in the unit's place, two places, and five numbers are left. Can't it be solved as 5C2?
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.
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?
Ramees said:
10 years ago
2, 3, 6, 7, 9 and 5 is must after two digits (because it must divisible by 5). Then we take 5 numbers (2, 3, 4, 7, 9) with 2 groups. Here we applies permutation (because 23 and 32 are different).
5p2 = 5!/(5-2)! = 5!/3! = 4*5 = 20.
Note that ab, ba are two different permutations but they represent the same combination.
5p2 = 5!/(5-2)! = 5!/3! = 4*5 = 20.
Note that ab, ba are two different permutations but they represent the same combination.
Joel said:
10 years ago
I need some help on how to come up with those digits?
Sara siddiqui said:
10 years ago
3p3*2C1 = 12.
Since we have four places and the no which is divisible by 4 must have 2 or 6 in the unit place, so 2 and 6 is consider for the unit place, since we are choosing between them so we use combination 2C1.
Now only 3 places are left with three digits 3, 7 and (2 or 6), now here we are arranging them so we use permutation so 3P3.
Since we have four places and the no which is divisible by 4 must have 2 or 6 in the unit place, so 2 and 6 is consider for the unit place, since we are choosing between them so we use combination 2C1.
Now only 3 places are left with three digits 3, 7 and (2 or 6), now here we are arranging them so we use permutation so 3P3.
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?
Eshwar VIRAT said:
1 decade ago
Any other questions like this in same pattern?
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