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:
89 comments Page 2 of 9.
Dema said:
8 years ago
How 5p2? Please explain.
(1)
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)
Ambika said:
9 years ago
Write all the possible number using the digit 7, 0, 6 repetition of digits is not allowed.
(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)
Monisha said:
9 years ago
@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.
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.
(1)
Pritesh said:
7 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)
Thanusha K said:
2 years ago
@ Vinay.
If you observe the digits given in question there is no '0'. So at units place, we won't consider '0'. And once that is 5. And each of them has 4 ways.
So, 5 * 4 = 20 ways.
If you observe the digits given in question there is no '0'. So at units place, we won't consider '0'. And once that is 5. And each of them has 4 ways.
So, 5 * 4 = 20 ways.
(1)
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.
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