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 8 of 9.
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.
Joel said:
10 years ago
I need some help on how to come up with those digits?
Sara siddiqui said:
1 decade 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.
Pranav said:
1 decade ago
How many four digit numbers divisible by 4 can be formed using the digits 2, 3, 6&7 the digits not being repeated?
Eshwar VIRAT said:
1 decade ago
Any other questions like this in same pattern?
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.
Xander said:
1 decade ago
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.
Basha said:
1 decade ago
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.
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.
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?
Poonam said:
1 decade ago
Hi friends.
The remaining 5 digits can be arranged in 5! ways or 5p2 ways.
The remaining 5 digits can be arranged in 5! ways or 5p2 ways.
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