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 7 of 9.
Riddhi said:
1 decade ago
Thanx rahul, your ans was awesome.
Ajay said:
1 decade ago
Please can you send the ans by divisible no by 3 but not by 5 with explanation?
Shree Hemalatha said:
1 decade ago
How the 4 digits are coming in 100's place?
Could you explain me that ?
Could you explain me that ?
Shanthi said:
1 decade ago
Please explain 1s, tens, hundreds place numbers in detail.
Rahul said:
1 decade ago
The answer can also be solved in another way:
To be divisible by 5, 5 will always be in units place. That leaves two numbers to be chosen for tens and hundreds place out of the remaining 5. Order matters here since 675 is different from 765. So, we will be choosing permutations which states that without repetition the number of permutations are n!/(n-r)! which is 5!/(5-2)! = 5!/3! = 5 x 4 = 20
To be divisible by 5, 5 will always be in units place. That leaves two numbers to be chosen for tens and hundreds place out of the remaining 5. Order matters here since 675 is different from 765. So, we will be choosing permutations which states that without repetition the number of permutations are n!/(n-r)! which is 5!/(5-2)! = 5!/3! = 5 x 4 = 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.
Nikita said:
2 decades ago
How are you considering 4 nos in hundreds place?.
Poonam said:
1 decade ago
Hi friends.
The remaining 5 digits can be arranged in 5! ways or 5p2 ways.
The remaining 5 digits can be arranged in 5! ways or 5p2 ways.
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.
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