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please help in permutation question

Expert replies
by sachin_yadav » Thu Dec 16, 2010 12:21 am
Hi All,

Please help in the following question.

A telephone company needs to create a set of 3-digit area codes. The company is entitled to use only digits 2, 4 and 5, which can be repeated. If the product of the digits in the area code must be even, how many different codes can be created ?


(A) 8

(B) 9

(C) 18

(D) 26

(E) 27


My answer is E which is incorrect and i am also aware of this mistake because i calculated this question without using a constraint. i.e Total number of codes = 3 * 3 * 3 = 27(each of the 3 digits can be repeated); however, the constraint says "product of the digits in the area code must be even".

The explanation says, " Number of codes yielding an even product = 27 - 1 = 26. So, answer is D (correct)

My question is :-

Why and what is the logic to subtract 1 from 27 ?

or

In order to get an even value, do we have to subtract 1 from 27 ?

Please advise..
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Source: — Problem Solving |

by shovan85 » Thu Dec 16, 2010 12:33 am
sachin_yadav wrote:Hi All,

Please help in the following question.

A telephone company
My answer is E which is incorrect and i am also aware of this mistake because i calculated this question without using a constraint. i.e Total number of codes = 3 * 3 * 3 = 27(each of the 3 digits can be repeated); however, the constraint says "product of the digits in the area code must be even".

The explanation says, " Number of codes yielding an even product = 27 - 1 = 26. So, answer is D (correct)

My question is :-

Why and what is the logic to subtract 1 from 27 ?

or

In order to get an even value, do we have to subtract 1 from 27 ?

Please advise..



Hi Sachin,

I really dont find any issue with your reasoning as you are very clear about what you are asked for and what you are looking for.
However, let me reiterate your reasoning...

Absolutely fine that you got 3*3*3 = 27.

Now see there is only one possibility of getting odd number when used all 3 digits that is 5*5*5 = 125.

You have to discard this odd number as q asks only about even product.

So discard 1 from 27 thus 27 - 1 = 26
If the problem is Easy Respect it, if the problem is tough Attack it
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by Night reader » Thu Dec 16, 2010 1:07 am
sachin_yadav wrote:Hi All,

Please help in the following question.

A telephone company needs to create a set of 3-digit area codes. The company is entitled to use only digits 2, 4 and 5, which can be repeated. If the product of the digits in the area code must be even, how many different codes can be created ?


(A) 8

(B) 9

(C) 18

(D) 26

(E) 27


My answer is E which is incorrect and i am also aware of this mistake because i calculated this question without using a constraint. i.e Total number of codes = 3 * 3 * 3 = 27(each of the 3 digits can be repeated); however, the constraint says "product of the digits in the area code must be even".

The explanation says, " Number of codes yielding an even product = 27 - 1 = 26. So, answer is D (correct)

My question is :-

Why and what is the logic to subtract 1 from 27 ?

or

In order to get an even value, do we have to subtract 1 from 27 ?

Please advise..
the only constraint here is 2,4,5 which are three digits to be used for our number

so perms will be 3*3*3 or 27

however, note 555 is not of the product even 5*5*5 is odd

so 27-1=26 perms
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by sachin_yadav » Thu Dec 16, 2010 11:49 am
Now see there is only one possibility of getting odd number when used all 3 digits that is 5*5*5 = 125.

You have to discard this odd number as q asks only about even product.
Thanks shovan and Night reader. After reviewing the explanation and trying the question again I have understood that any other code containing at least one 2 or 4 will always yield an even product and the product of the 3 numbers will be odd only if all the digits are odd and in this case 555 violates the constraint.

Thank you once again.
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