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 2 of 9.
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.
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)
Himabindu said:
1 decade ago
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.
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.
Vivek Kumar said:
1 decade ago
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)!
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)!
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.
Vimala said:
1 decade ago
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.
so 5*4*3=60 ways.
I hope that it is right. If it is wrong please help me.
Sunthar said:
7 years ago
(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.
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.
(1)
Pritesh said:
6 years ago
1. 5 should come at unit place.
2. Remaining two i.e. tens and hundred places should be filled by remaining five nos. So we need to first select two nos. Among five i.e 5C2.
3. Arrange those selected two nos = 2!.
4. So, the required Answer = 5C2*2!=20.
2. Remaining two i.e. tens and hundred places should be filled by remaining five nos. So we need to first select two nos. Among five i.e 5C2.
3. Arrange those selected two nos = 2!.
4. So, the required Answer = 5C2*2!=20.
(1)
Santiswarup mishra said:
1 decade ago
(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
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
Dhiraj kumar said:
1 decade ago
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
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
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