Aptitude - Permutation and Combination - Discussion

Discussion :: 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?

[A]. 5
[B]. 10
[C]. 15
[D]. 20

Answer: Option D

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.


Swati Patil said: (Jul 23, 2010)  
Please can you send the ans by divisible no by 3 but not by 5 with explanation?

Nikita said: (Sep 3, 2010)  
How are you considering 4 nos in hundreds place?.

Hari said: (Sep 10, 2010)  
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.

Rahul said: (Oct 28, 2010)  
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

Shanthi said: (Dec 12, 2010)  
Please explain 1s, tens, hundreds place numbers in detail.

Shree Hemalatha said: (Feb 12, 2011)  
How the 4 digits are coming in 100's place?

Could you explain me that ?

Ajay said: (Apr 23, 2011)  
Please can you send the ans by divisible no by 3 but not by 5 with explanation?

Ashu said: (Jun 29, 2011)  
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)

Riddhi said: (Jul 24, 2011)  
Thanx rahul, your ans was awesome.

Karthika said: (Aug 4, 2011)  
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!

Dhiraj Kumar said: (Aug 10, 2011)  
As the requried condition is XY5.

Since 5 is fixe so we have to select two nos (X & Y) out of remaining five nos i.e,
5C2 & Two nos X&Y arrange in 2! Ways.
=> Total no. Way= 5C2* 2!

= (10)*2 = 20

Raj said: (Aug 12, 2011)  
Can any one explain what is the procedure if the digits are repeated for the above problem?

Deepu said: (Aug 24, 2011)  
Karthikas answer is the most easy to understand and generic concept. I recommend to follow that method.

Deva said: (Aug 27, 2011)  
Thanks karthika and ashu.

Raj said: (Oct 7, 2011)  
Well its very simple to think as my frnds says that we shd find 3 values and three place in three digit it means unit place tens place and hundreds place as they told the no. is divisible by 5 so the no. at unit place is 5

So now we are having only 2 place remaning i:e tens and hundreds so while using factorial permutation npr n!/(n-r)! were n + number we have i;e 5 r= the sum of places i;e 2 so 5!/(5-2)!=5!/3!=5*4*3*2*1/3*2*1=5*4 =20

Santiswarup Mishra said: (Jan 10, 2012)  
(mark that none of the digits are repeated)
We must use 5 in units(bcz to b divisible wid 5 only/)
then we have 4 other nos. to use in tens then
again to avoid repetation of no.s we should use 3 nos in the hundreds place.
so according 2 me ans is 1*4*3=12

Vivek Kumar said: (Jan 11, 2012)  
Can also be solved as :---

Units place is fixed so only one method for it.

Now for tens and hundreds place....we can use Permutations as here ordering of digits matters.
As 5 is already reserved at units place so we are left with two choices..
So it becomes :

5P2 * 1

5!
= ------ * 1 = 5x4 = 20 Ans..
(5-2)!

Hanipoo23 said: (Feb 23, 2012)  
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.

Anurag said: (Mar 6, 2012)  
Hi..
my ques is that why did we not do 5! x 4! x 1!
as per which it should be 5*4*3*2*4*3*2*1
and not 5*4*1.........

and secondly why did we go for permutation over combination here??

Chandra said: (May 29, 2012)  
n!/ (n-r) ! is the best formula and easy way to get answer.

Ayodele K said: (Oct 9, 2012)  
If the three digits must be divisible by 5 and no digit is to be repeated, therefore, it's safe to reserve a spot for 5 and then have 2,3,6,7,9 combine the first two digits (XY5),
Permutation of the first two digits gives 5C2
5!/3!

23,26,27,29
32,36,37,39
62,63,67,69
72,73,76,79
92,93,96,97

These outcomes combine with 5 each. 5P3*1 resulting in 20 outcomes.

Remember, no repetition, so there

Suganya said: (Oct 11, 2012)  
Please explain the 1s 10s &100 place in number detaily.

Sujit said: (Nov 27, 2012)  
@Suganya.

In number 456.

6 is in Unit Place.
5 is in 10th place.
4 is in 100th place.

Hope you get this :-).

Sheraz said: (Dec 28, 2012)  
How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated in a number?

Shashank said: (Mar 20, 2013)  
@Raj.

Given Digits: 2,3,5,6,7,9

If the digits are repeated then

Let 3 digit no. be XYZ.

Divisible by 5 must have 5 at Z place or unit place i:e XY5. that is only one way to fill Z place.

Now Y place can be filled by any of 6 digits i:e 2,3,5,6,7,9
in 6C1 = 6 ways.

Also X place can be filled by any of 6 digits i:e 2,3,5,6,7,9
in 6C1 = 6 ways.

So total ways = 6*6*1 = 36 ways.

I think its clear to you know. :).

Gold said: (Jun 5, 2013)  
Since it says that we have to look for numbers that are divisible by 5 among 2,3,5,6,7,9 so the unit digit has to be 5.
__ __ _5_
Since we don't know which numbers are going to be on the first and second (hundredth and tenth) we have to exclude 5 and think about all the cases that can be above.

2,3,6,7,9 are left except 5. So 5 numbers can go into the first digit which is hundredth, and since we used 2 numbers so far, we subtract 2 from the total number of the numbers on the top.

6-2 = 4 (left).

So in the second digit, only 4 numbers can go in.
Hence, 5*4*1 = 20 (answer)

Athul said: (Oct 1, 2013)  
5 has to come in the last digit. So its fixed. Now we are concerned only about the first two digits. We can choose from remaining five no's in 5C2 ways. And this can be arranged in 2! ways. So ans is 5C2*2! = 20.

Vimala said: (Nov 6, 2013)  
The correct answer is 60 ways. The logic behind this is, we know that when a number is divisible by 3 the sum of the digit should be divisible by 3. In the given number, only 2,3,6,7,9 are divisible by 3 while adding. 2+3+6+7+9=27,
so 5*4*3=60 ways.

I hope that it is right. If it is wrong please help me.

Ankit Sharma said: (Dec 27, 2013)  
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?

Dushyant Verma said: (Jan 27, 2014)  
The 3 digit no should be divisible by 5 so it must the 5 is at the unit place then the 3 digit no will be ---- xy5.

Now the two no. (x and y ) are taking from five ( 2,3,6,7,9).

So 5p2 = 5*4 = 20.

Preethi said: (Apr 10, 2014)  
n!/(n-r) !.r!, is the best way to solve the problem.

Surbhi Srivastava said: (Apr 14, 2015)  
The greatest 4 digit number which is only divisible by 5.

Siva said: (Apr 30, 2015)  
Lets see the detail combination divisible by 5 are 235, 265, 275, 295.

Similarly for other numbers. Hence there are 5 number and 4 combination for each number.

Poonam said: (May 30, 2015)  
Hi friends.

The remaining 5 digits can be arranged in 5! ways or 5p2 ways.

Tanzeem said: (May 31, 2015)  
The question is incomplete and confusing, it has to mention that none of the digit to be repeated in the number. If we consider the condition then 5 is repeated in every number.

Basha said: (Jun 29, 2015)  
The 3-digit no. may be" XY5".

At 100th position 5 is fix because it divisible by 5.

Hence 5 occur only once.

Next 10th place Y it may be 2, 3, 6, 7, 9.

Any one of the no. Can occur hence it is 5 i.e. any one can occur.

Next 1st position X can be filled with remaining 4 no. Because of no repetition i.e the chance is 4.

The 3-digits is -1*5*4 = 20.

Xander said: (Jul 29, 2015)  
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.

Himabindu said: (Aug 20, 2015)  
Let us take the three places be_ _ _, to be divisible by 5 the last place should be 5, so we take 5 in the units place_ _ 5.

Now in the tens place we take one of the remaining five numbers so, it becomes _ 5C1 5, now the hundreds place will be full filed by one of the left over four numbers, so 4C1.

Hence 4C1*5C1*1C1.

= 4*5*1 = 20.

Eshwar Virat said: (Aug 30, 2015)  
Any other questions like this in same pattern?

Pranav said: (Sep 4, 2015)  
How many four digit numbers divisible by 4 can be formed using the digits 2, 3, 6&7 the digits not being repeated?

Sara Siddiqui said: (Nov 13, 2015)  
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.

Joel said: (Jan 21, 2016)  
I need some help on how to come up with those digits?

Ramees said: (Feb 4, 2016)  
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.

Java Eucy said: (Mar 28, 2016)  
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?

Dharang said: (Jul 7, 2016)  
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.

Shivam said: (Jul 8, 2016)  
Say number 5 is placed in the unit's place, two places, and five numbers are left. Can't it be solved as 5C2?

Akhil said: (Jul 13, 2016)  
Thanks @Rahul! good explanation.

Pravas said: (Jul 22, 2016)  
Shortcut way to solve this problem :-6C3 => 6!/3!
= 6 * 5 * 4/3 * 2 * 1.
= 20.

Shivam said: (Aug 16, 2016)  
Why can't we write 5c2? There are two places and 5 numbers.

Chandrashekar said: (Sep 25, 2016)  
Here, the answer is 20.

By using the fundamental principle of counting ie, if one activity can be done in 'm' num of different ways and the second activity can be done in 'n' number of different ways and these two activities can be done in MxN different ways.

For example, how mane 3 digit number can be formed By using the numbers 2, 3, 5, 8.without repeating.
Soln:-
Let the numbers are 2, 3, 5, 8 and u all know 3 digit number means it must have unit tens and hundred places.
So, now you just Make three column like H T U.
2 3 4 ie in unit place you can fill any of the 4 numbers means 4 different ways and remaining 2 place that is tens and hundred and in the tens place, you can fill in 3 different ways because repetition not allowed.
If once you used the digit you cannot use again. And next 100 places you can fill in 2 different ways.

Therefore total no of ways = 2 x 3 x 4.
= 24ways.
I Hope you understood.

Chandrashekar Pa said: (Sep 25, 2016)  
Divisibility rule 3:

If the sum of the digit is multiple of 3 then that number is divisible by 3.

Ex: 123 is divisible by 3. 1 + 2 + 3 = 6 here 6 is multiple of 3 therefore 123 is divisible by 3.

Ayush Jha said: (Sep 27, 2016)  
How many 5-digited numbers can be formed by using 0, 1, 2, 3, 4, 5, 6 which is divisible by 7 when repetition is allowed?

Please help me to get the answer.

Kalai said: (Oct 1, 2016)  
In the third explanation I didn't understand please explain it.

Peter said: (Oct 12, 2016)  
Thanks a lot for the explanation. It is easy now.

Venkat said: (Oct 17, 2016)  
Can any one help me on this one? How many 5 digit numbers can be formed from the digits 0 to 9, so that odd digits are occupied only in even position using one digit only once.

Manpreet said: (Nov 13, 2016)  
How 4 digits are remaining?

Otieno Bonface Omondi said: (Nov 20, 2016)  
I am not understanding, what has been done. Can someone elaborate the whole thing in a simpler mathematical language? Please.

Vijay015125 said: (Nov 21, 2016)  
On units place 5 is fixed.
On 10's place remaining 5 digits one is selected = 5.

At finally 100's place remaining 4digits will be there so 4.
RESULT = 4 * 5 * 1 = 20.

Ashutosh said: (Nov 29, 2016)  
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.

Justin said: (Dec 5, 2016)  
The best way is,

6c3 = 20.

Karthick said: (Jan 23, 2017)  
How many 4 digit no can be formed with digits 1, 2, 3,4,5 which are divisible by 4 and digits not repeated?

a)144 b) 168 c) 182 d) none

Please tell me the correct answer.

Manisha said: (Jan 31, 2017)  
Nice answer@Justin.

Monisha said: (Feb 21, 2017)  
@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.

Joe said: (Mar 8, 2017)  
Why isn't it just 5C2, from the 5 numbers that arent 5 to go in tens and hundreds, and picking two?

Vignesh said: (Apr 27, 2017)  
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: (Jun 15, 2017)  
Write all the possible number using the digit 7, 0, 6 repetition of digits is not allowed.

Xiya said: (Jul 11, 2017)  
Can anyone say is 20 the right answer?

Because the question asked is number divisible by 5 and also not repeated?

Dema said: (Aug 4, 2017)  
How 5p2? Please explain.

Pranav Khurud said: (Aug 4, 2017)  
If question is like, How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7, 0, and 9, which are divisible by 5 and none of the digits is repeated?, In this situation unit place may be 0 or 5.

Shantanu said: (Sep 28, 2017)  
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?

Chals said: (Oct 5, 2017)  
Can anyone answer for this?

How many can 5 digit number divisible by 4 be formed using the digits 1, 2, 3, 4, 5, 6 no digits being repeated in the numbers?

Raoakc said: (Nov 19, 2017)  
Can't we do it by combination?

Brown said: (Jul 25, 2018)  
What if the series includes 0, ie 2 3 5 6 7 9 0?

Sunthar said: (Aug 9, 2018)  
(23679) 5.

The format should be like XY5.

Since 5 is fixed in Once position so that only outcome number will be divisible by 5.

Out of 3 positions, 2 only left.

So that 2 numbers from 5 numbers (2, 3, 6, 7, 9) can occupy the 2 spots.

(i.e) 5p2 = 5!/(5-2)! = 120/6 = 20.

5 remaining numbers. 2 spots.

Pooja said: (Nov 14, 2018)  
Well explained. Thanks.

Shaurya said: (Mar 24, 2019)  
To divide with the no 5, no must have 5 at its unit place.

So 5 is fixed at a unit place now there are 5 digits that are left and we have to choose only 2 of them so 5p2.

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