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How many 4 digit even numbers such that the digits are 2, 4, 5 and 8?

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How many 4 digit even numbers such that the digits are 2, 4, 5 and 8? [#permalink]

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How many 4 digit even numbers such that the digits are 2, 4, 5 and 8?

A) 4*4*4*4
B) 3*4*4*4
C) 3*3*3*3
D) 3*3*2*1
E) 3*2*2*1


Source => Kaplan.
I have mentioned my doubt in comments.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 14 Aug 2017, 00:43, edited 1 time in total.
Renamed the topic and edited the question.

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Re: How many 4 digit even numbers such that the digits are 2, 4, 5 and 8? [#permalink]

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New post 13 Aug 2017, 22:19
Here is my Question => Without telling us the repetition is allowed or not how can we answer.
In the current state -> It should be 4*4*4*3 and not otherwise.

What am I missing ?
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Re: How many 4 digit even numbers such that the digits are 2, 4, 5 and 8? [#permalink]

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New post 13 Aug 2017, 22:27
stonecold wrote:
Here is my Question => Without telling us the repetition is allowed or not how can we answer.
In the current state -> It should be 4*4*4*3 and not otherwise.

What am I missing ?

Hi stonecold,
I will also approach in the same way as have you.

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Re: How many 4 digit even numbers such that the digits are 2, 4, 5 and 8? [#permalink]

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New post 13 Aug 2017, 22:30
My reasoning for this approach I stated earlier is that at the units place we cannot have 5 as it will result in an odd number. So rest of the places i.e thousands , hundreds and tens digits can be filled in 4 ways each.

I am not sure if so am really missing something.

Pls help.
Thanks in advance?

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Re: How many 4 digit even numbers such that the digits are 2, 4, 5 and 8? [#permalink]

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New post 13 Aug 2017, 22:40
stonecold wrote:
Here is my Question => Without telling us the repetition is allowed or not how can we answer.
In the current state -> It should be 4*4*4*3 and not otherwise.

What am I missing ?


Hi,

You are right. Answer should be 4*4*4*3 else one has to mention that repetition is not allowed.

The OA is based on the without repetition case.


Thanks.

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Re: How many 4 digit even numbers such that the digits are 2, 4, 5 and 8? [#permalink]

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New post 13 Aug 2017, 22:43
Update ->
Received an email from Kaplan's team -->



Edit: Thank you for reaching out to us with your question. When they are asking you for a 4-digit number, it is a permutations question because it asks for the number of 4-digit numbers "such that the digits are 2, 4, 5, and 8". The phrasing of the question is tricky. It's not how many even 4-digit numbers are there made up of 2, 4, 5, and 8 (where you could have 4 eights to make an even 4-digit number). It says the digits ARE 2, 4, 5 and 8. So the number can only have those 4 digits and only one of each of them. So the digits are distinct.



So what they are saying is -> If the question says made up of "2, 4, 5, and 8" that means every digit must be used.
I am not sure if that is the right approach but yeah I get it now.


Any other views on this one ?


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Re: How many 4 digit even numbers such that the digits are 2, 4, 5 and 8? [#permalink]

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New post 13 Aug 2017, 23:32
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stonecold wrote:
Update ->
Received an email from Kaplan's team -->



Edit: Thank you for reaching out to us with your question. When they are asking you for a 4-digit number, it is a permutations question because it asks for the number of 4-digit numbers "such that the digits are 2, 4, 5, and 8". The phrasing of the question is tricky. It's not how many even 4-digit numbers are there made up of 2, 4, 5, and 8 (where you could have 4 eights to make an even 4-digit number). It says the digits ARE 2, 4, 5 and 8. So the number can only have those 4 digits and only one of each of them. So the digits are distinct.



So what they are saying is -> If the question says made up of "2, 4, 5, and 8" that means every digit must be used.
I am not sure if that is the right approach but yeah I get it now.


Any other views on this one ?



I think their explanation makes sense, because it is saying that the digits are 2,4,5 and 8
For it to be without repetition, it must have been 2,4,5 or 8!

There are 3 possibilities where the digit can be : 1st,2nd or 3rd digit.
The remaining even numbers can be written in 3*2*1 ways
So, it must be 3*3*2*1 or 18 ways(Option D)
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Re: How many 4 digit even numbers such that the digits are 2, 4, 5 and 8? [#permalink]

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New post 14 Aug 2017, 00:22
I think the question should say " How many even four digit number can be made using all the digits from 2, 4, 5 & 8, without repetition?"
Then its answer would be 3 X 2 X 1 X 3 = 18.

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Re: How many 4 digit even numbers such that the digits are 2, 4, 5 and 8? [#permalink]

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New post 14 Aug 2017, 01:35
My opinion: In case that any quant question is confused, the reason is either the editor was lazy to type something or that question is poor quality. Hence, no need to waste time on that question.
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Re: How many 4 digit even numbers such that the digits are 2, 4, 5 and 8? [#permalink]

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New post 25 Aug 2017, 10:59
I think the logic is the following
four digits number xxxx should be even
so starting from the last digit it can take only 3 possible values 2, 4 and 8
for the second last it's also 3 as we can use 5
then 2 and finally 1

3*3*2*1 permutation as the question asks HOW Many possible unique numbers

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Re: How many 4 digit even numbers such that the digits are 2, 4, 5 and 8?   [#permalink] 25 Aug 2017, 10:59
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