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How many 3 digit even numbers can be formed with the digits 0,1,2,3,4,5 if no digit is repeated in any number?

A. 48
B. 50
C. 52
D. 54
E. 100

amit482001 wrote:
How many 3 digit even numbers can be formed with the digits 0,1,2,3,4,5 if no digit is repeated in any number?

A. 48
B. 50
C. 52
D. 54
E. 100

hi amit...same question has been raised by khalida in another thread...
answer to this question is 52...i shall post up the detailed video soultion tomorrow morning..

keep going...

hi,
thanks for the reply.
but is there any genreal trend for doing these kind of problems, because i didnt understand this kind of problem when we were asked to get a 5 digit number in the same thread.
what i mean is i understood the video, but i failed to understand how to go about doing it, could you please tell me if there is any general method or trick that i can go about doing these problems

amit482001 wrote:
hi,
thanks for the reply.
but is there any genreal trend for doing these kind of problems, because i didnt understand this kind of problem when we were asked to get a 5 digit number in the same thread.
what i mean is i understood the video, but i failed to understand how to go about doing it, could you please tell me if there is any general method or trick that i can go about doing these problems

Hi Amit,

You have a very valid question, but i shall appreciate if you can post your query in the same relevant thread i.e post your query in the exisiting thread of number theory, so others can also understand the context of the problem.

Approach for these kind of problems...

Always begin with number of ways for most restricted position.

eg. How many 3 digit even numbers can be formed with the digits 0,1,2,3,4,5 if no digit is repeated?

There are 2 restrictions:

1) last digit should be even
2) First digit should not be zero.

To meet last digit requirement, there are again 2 possibilities :
1) last digit could be zero, since 0 is even number
OR
2) last digit is any other even number but zero from available digits.

then starting from the restricted position, count the no of digits that can occupy every position.

i.e evaluate total number of ways in each of the above 2 conditions and add them to get final result.

To understand even better, refer to video solution in Number theory thread..

another eg. How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?

Now, in this case, number should be divisible by 3. And divisibility test of 3 says that if sum of digits is divisible by 3, then number is also divisible by 3.

Hence consider combination of 5 digits, whose sum is multiple of 3.

There are only 2 possibilities . ie digits could be:
0,1,2,4,5
OR
1,2,3,4,5

in the second combination there is no restriction in postion of nos. ie any digit can occupy any position.

For the first combination of digits, first place is restricted , since 0 cannot occupy first position.

hence you count the possible digits which can occupy every position and multiply them.

Hope this solves your query.

Keep going..