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 3 of 9.
Athul said:
1 decade ago
5 has to come in the last digit. So its fixed. Now we are concerned only about the first two digits. We can choose from remaining five no's in 5C2 ways. And this can be arranged in 2! ways. So ans is 5C2*2! = 20.
Pranav Khurud said:
8 years ago
If question is like, How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7, 0, and 9, which are divisible by 5 and none of the digits is repeated?, In this situation unit place may be 0 or 5.
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.
Vignesh said:
8 years ago
Please explain me, what would happen if another digit (0) is added to the series, i.e., 2,3,6,7,9,5 and 0 and the digits are to arranged without repetition such that the 3 digit number is divisible by 5?
Chandrashekar pa said:
9 years ago
Divisibility rule 3:
If the sum of the digit is multiple of 3 then that number is divisible by 3.
Ex: 123 is divisible by 3. 1 + 2 + 3 = 6 here 6 is multiple of 3 therefore 123 is divisible by 3.
If the sum of the digit is multiple of 3 then that number is divisible by 3.
Ex: 123 is divisible by 3. 1 + 2 + 3 = 6 here 6 is multiple of 3 therefore 123 is divisible by 3.
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.
Anurag said:
1 decade ago
Hi..
my ques is that why did we not do 5! x 4! x 1!
as per which it should be 5*4*3*2*4*3*2*1
and not 5*4*1.........
and secondly why did we go for permutation over combination here??
my ques is that why did we not do 5! x 4! x 1!
as per which it should be 5*4*3*2*4*3*2*1
and not 5*4*1.........
and secondly why did we go for permutation over combination here??
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)
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.
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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