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 1 of 9.
Swati patil said:
2 decades ago
Please can you send the ans by divisible no by 3 but not by 5 with explanation?
Nikita said:
2 decades ago
How are you considering 4 nos in hundreds place?.
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.
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
Shanthi said:
1 decade ago
Please explain 1s, tens, hundreds place numbers in detail.
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 ?
Ajay said:
1 decade ago
Please can you send the ans by divisible no by 3 but not by 5 with explanation?
Ashu said:
1 decade ago
5 is at unit place.. so the combinations for 10's place are
25, 35, 65,75,95
for 100's place wid no digits r repeated den combinations will b
25 325 625 725 925
35 235 635 725 925
65 ...
75 ...
95 ...
like wise last two digits r fixed.... nd only first is varied..
so the combinations r
4(no. wich r in rows0* 5(No. which r in columns)
25, 35, 65,75,95
for 100's place wid no digits r repeated den combinations will b
25 325 625 725 925
35 235 635 725 925
65 ...
75 ...
95 ...
like wise last two digits r fixed.... nd only first is varied..
so the combinations r
4(no. wich r in rows0* 5(No. which r in columns)
(1)
Riddhi said:
1 decade ago
Thanx rahul, your ans was awesome.
Karthika said:
1 decade ago
It explained in a bit confusing way...
This calculation can be explained in a simpler way....
As the 3 digit number must be a multiple of 5, it must have 5 in its units place
Ex: xy5. X and Y can be any of the remaining 5 digits(2,3,6,7,9).
These remaining 5 digits can have 5p2 permutations ( P- because the arrangement is important).
That gives 20 possible arrangements of the 5 digits.
The number of possible permutations of the digit '5' is 5p5 = 1.
Therefore 5p2 and 5p5 ( 20 * 1) gives 20.
Here's the answer!
This calculation can be explained in a simpler way....
As the 3 digit number must be a multiple of 5, it must have 5 in its units place
Ex: xy5. X and Y can be any of the remaining 5 digits(2,3,6,7,9).
These remaining 5 digits can have 5p2 permutations ( P- because the arrangement is important).
That gives 20 possible arrangements of the 5 digits.
The number of possible permutations of the digit '5' is 5p5 = 1.
Therefore 5p2 and 5p5 ( 20 * 1) gives 20.
Here's the answer!
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers