Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 7)

Permutation and Combination

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 8 of 9.

Preethi said: 1 decade ago

n!/(n-r) !.r!, is the best way to solve the problem.

Peter said: 9 years ago

Thanks a lot for the explanation. It is easy now.

Nikita said: 2 decades ago

How are you considering 4 nos in hundreds place?.

Vinay said: 6 years ago

0 may also come in unit place, please explain me.

Brown said: 7 years ago

What if the series includes 0, ie 2 3 5 6 7 9 0?

Sarvani said: 6 years ago

Please explain it clearly. I didn't understand.

(2)

Eshwar VIRAT said: 1 decade ago

Any other questions like this in same pattern?

Riddhi said: 1 decade ago

Thanx rahul, your ans was awesome.

Akhil said: 9 years ago

Thanks @Rahul! good explanation.

Raoakc said: 8 years ago

Can't we do it by combination?

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