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 6 of 9.
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.
Tanzeem said:
1 decade ago
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.
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.
Siva said:
1 decade ago
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.
Similarly for other numbers. Hence there are 5 number and 4 combination for each number.
Surbhi srivastava said:
1 decade ago
The greatest 4 digit number which is only divisible by 5.
Preethi said:
1 decade ago
n!/(n-r) !.r!, is the best way to solve the problem.
Dushyant Verma said:
1 decade ago
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.
Now the two no. (x and y ) are taking from five ( 2,3,6,7,9).
So 5p2 = 5*4 = 20.
Ankit Sharma said:
1 decade ago
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?
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?
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