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.
CHALS said:
8 years ago
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?
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?
(1)
Manpreet said:
9 years ago
How 4 digits are remaining?
(1)
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)
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)
Otieno Bonface Omondi said:
9 years ago
I am not understanding, what has been done. Can someone elaborate the whole thing in a simpler mathematical language? Please.
Vijay015125 said:
9 years ago
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.
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:
9 years ago
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.
Solve this and find the solution.
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?
Venkat said:
9 years ago
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.
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