telephone extensions

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telephone extensions [#permalink]

08 Jun 2009, 06:13

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If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of four-digit extensions that the company can have?

(A) 4 (B) 6 (C) 12

(D) 16 (E) 24

[spoiler]OA:(c)[/spoiler]

The wording of the problem itself is not clear.I could'nt figure out what we are supposed to find in the first place?Nevertheless I answered 4!=24 which is wrong.Can someone please shed some light?
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Re: telephone extensions [#permalink]

08 Jun 2009, 13:24

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it should be 12 ---

digit 1 * digit 2 * digit 3 * digit 4

digit 4 -- last digit is having 2 option only even.
digit 3 -- left out 3
digit 2 has left out 2
digit 1 has left out 1.

multiply all ways u get 1*2*3*2 = 12....

please confirm.

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Re: telephone extensions [#permalink]

09 Jun 2009, 16:04

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The answer is C.

You are on the right track..the total number of permutations is 4! = 24

However there is a restriction that all extensions should be even i.e. the last digit of each extension should be an even number i.e. the last digit should be either 2 or 6. As there are 2 even digits and two odd digits. Half of the permutations should be even and the remaining odd.

Divide 24/2 results into 12

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Re: telephone extensions [#permalink]

09 Jun 2009, 16:29

one more way....
let digit4 = 6, the 3! is the way to arrange rest 3 digit, so total 6 ways
let digit4 = 2, the 3! is the way to arrange rest 3 digit, so total 6 ways

so total 12 ways (C)

Re: telephone extensions [#permalink] 09 Jun 2009, 16:29

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