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 9 of 9.
Manpreet said:
9 years ago
How 4 digits are remaining?
(1)
Justin said:
9 years ago
The best way is,
6c3 = 20.
6c3 = 20.
Deva said:
1 decade ago
Thanks karthika and ashu.
Dema said:
8 years ago
How 5p2? Please explain.
Pooja said:
7 years ago
Well explained. Thanks.
Akash said:
5 years ago
It is 5C2 * 2! = 20.
(8)
Manisha said:
9 years ago
Nice answer@Justin.
Alok said:
6 years ago
Thanks @Sujit.
(2)
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