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?
5
10
15
20
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.

Sujit said:   1 decade ago
@Suganya.

In number 456.

6 is in Unit Place.
5 is in 10th place.
4 is in 100th place.

Hope you get this :-).

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?

Deepu said:   1 decade ago
Karthikas answer is the most easy to understand and generic concept. I recommend to follow that method.

Joe said:   9 years ago
Why isn't it just 5C2, from the 5 numbers that arent 5 to go in tens and hundreds, and picking two?

Raj said:   1 decade ago
Can any one explain what is the procedure if the digits are repeated for the above problem?

Ambika said:   8 years ago
Write all the possible number using the digit 7, 0, 6 repetition of digits is not allowed.

Pravas said:   9 years ago
Shortcut way to solve this problem :-6C3 => 6!/3!
= 6 * 5 * 4/3 * 2 * 1.
= 20.

Ajay said:   1 decade ago
Please can you send the ans by divisible no by 3 but not by 5 with explanation?

Swati patil said:   2 decades ago
Please can you send the ans by divisible no by 3 but not by 5 with explanation?

Dinesh s said:   6 years ago
H T O
_ _ 5.
So two chance only.
Therefore ans 5C2 = 5 x 4 = 20 simple.
(8)


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