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If all of the telephone extensions in a certain company must

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If all of the telephone extensions in a certain company must [#permalink] New post 22 Jan 2005, 13:03
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83% (02:40) correct 17% (00:00) wrong based on 6 sessions
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
[Reveal] Spoiler: OA
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If all of the telephone extensions in a certain company must [#permalink] New post 22 Jan 2005, 13:16
let the 4 digits be

the possible ways are

A B C D

1 * 2 *3 * 2 = 12

total 12 ways

the logic is simple

4 digits are there so the units shud be 2 or 6 tht can be counted in 2 ways

for the next digit there are 3 left so 3 ways the next digit 2 left 2 ways last one can be chosen in 1 way for the lefot over digit

so 1* 2* 3* 2 = 12


P.S. but also if repetetions are allowed then it will be

4* 4 * 4* 2 = 128
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Re: If all of the telephone extensions in a certain company must [#permalink] New post 23 Jan 2005, 16:27
Why is it not 48? (I know it's not an option)
Like 4*3*2*2?

Rationale:

We are combining 4 digit extensions which must be even numbers. ABCD i.e., D but be either 2 or 6 from the given digits 1, 2, 3, 6.
Now we are left with the first 3 spaces which can be filled with any of these 4 given digits 1, 2, 3 or 6 which implies 4*3*2.
I stand corrected though. :wink:
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Re: If all of the telephone extensions in a certain company must [#permalink] New post 23 Jan 2005, 23:44
Folaa3 wrote:
Now we are left with the first 3 spaces which can be filled with any of these 4 given digits 1, 2, 3 or 6 which implies 4*3*2.

Each number must use all of the four numbers. So after you pick the unit digit, you only have three numbers to pick, not four.
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Re: If all of the telephone extensions in a certain company must [#permalink] New post 24 Jan 2005, 08:39
nice question,

*** 1st, the number can be expressed as "ABC2" (ending in 2)

so, there are 3*2*1*1 = 6 ways to express the number using 1,2,3,6


*** 2nd, the number can be expressed as "ABC6" (ending in 6)

so, there are 3*2*1*1 = 6 more ways to express the number using 1,2,3,6


therefore: 6 + 6 = 12
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Re: PS telephones numbers [#permalink] New post 09 Jun 2005, 01:47
mandy wrote:
:) .hello
Does anyone knows how to approach this one
thanks PS
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


The four digits available are 1,2,3 and 6
as the extensions can only be even numbers, the last digit has to be either 2 or 6
If it is 2...it leaves 1,3 and 6 for the first three digits...these can be arranged in 3! ways
If the last digit is 6...it leaves 1,2 and 3 for the first three digits...these can be arranged in 3! ways

Thus answer is 2*3! = 12
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Re: PS telephones numbers [#permalink] New post 09 Jun 2005, 08:24
mandy wrote:
:) .hello
Does anyone knows how to approach this one
thanks PS
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


4!/2 ,

The total no. extensions possible is 4!.
Of which, equal no. extns can be odd and even.
Hence the ones which are even is 4!/2.

HMTG.
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Re: PS - 1000PS [#permalink] New post 20 Apr 2007, 21:40
[quote="surbab"]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

Please explain the answer.[/quote]

(C)

Let's see... the conditions are that:
- 4-digits telephone numbers
- the only digits we can consider are 1,2,3,6
- the telephone extensions must be even numbers, so they will finish in 2 or 6
- ALL the digits (1,2,3,6) should be used in the same telephone number

total # of even numbers = total # with final digit 2 + total # with final digit 6
= 3*2*1 (the 4th digit is nr. 2) + 3*2*1 (the 4th digit is nr. 6)
=12

The easiest way to solve this problem is to think the telephone number digit by digit.
e.g.
4th digit: it must be even, so lets assume that it finishes in 2 -> 1
1st digit: we have 3 options (1 or 3 or 6) -> 3
2nd digit: we have 2 options (1,3 or 1,6 or 3,6) -> 2
3rd digit: we have only 1 nr left (1 or 3 or 6) -> 1
When you multiply all the numbers calculated per digit you get the number of possibilities: 1*3*2*1 = 6
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 [#permalink] New post 20 Apr 2007, 22:07
Since the phone number must be even, the unit's digit can be either 2 or 6.

When the unit's digit is 2 --> number of possibilities is 3! = 6
When the unit's digit is 6 --> number of possibilities is 3! = 6

Largest number of extensions = 6 + 6 = 12

Answer: C
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 [#permalink] New post 21 Apr 2007, 05:19
For the extension to be even the last digit should be 2 or 6.
Number of combinations with 2 be the last digit = 6
Number of combinations with 6 be the last digit = 6
Total possible cases = 12

Hence answer is 'C'
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 [#permalink] New post 21 Apr 2007, 05:26
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

Please explain the answer.

All possibilities for the combination of 4 digits is 4!=4x3x2x1= 24. Since we only need even numbers an we do have 2 even and 2 odd numbers it is 50% or 12!

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Re: telephone extensions [#permalink] New post 08 Jun 2009, 12:24
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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] New post 09 Jun 2009, 15:04
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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] New post 09 Jun 2009, 15:29
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)
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Re: A math problem [#permalink] New post 24 Jul 2011, 05:55
tracyyahoo wrote:
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, 2,3,6, what is the greatest number of four-digit extensions that the company can have?

a)4
b)6
c)12
d)16
e)24

pls help


The number has to be even hence the 4 digit number should end with either 2 or 6. Secondly the question states that all the numbers must be used hence repetitions are not allowed.

Hence: We can have _ _ _ 2 or _ _ _ 6. So now we need to only fill 3 places with digits 1,6,3 or 1,2,3 respectively
Hence 3!*2 = 12.

C
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Re: A math problem [#permalink] New post 25 Jul 2011, 02:20
Could u explain more details?




uote="Sudhanshuacharya"]
tracyyahoo wrote:
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, 2,3,6, what is the greatest number of four-digit extensions that the company can have?

a)4
b)6
c)12
d)16
e)24

pls help


The number has to be even hence the 4 digit number should end with either 2 or 6. Secondly the question states that all the numbers must be used hence repetitions are not allowed.

Hence: We can have _ _ _ 2 or _ _ _ 6. So now we need to only fill 3 places with digits 1,6,3 or 1,2,3 respectively
Hence 3!*2 = 12.

C[/quote]

Posted from my mobile device Image
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Re: A math problem [#permalink] New post 25 Jul 2011, 03:08
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, 2,3,6, what is the greatest number of four-digit extensions that the company can have?

a)4
b)6
c)12
d)16
e)24

The number for the telephone extension has to be even and 4 digit. Hence the 4 digit number should end with either 2 or 6.
Also, the question states that all the numbers must be used because we are talking about greatest number of four-digit extensions.

So, we can have _ _ _ 2 or _ _ _ 6 as four digit even extensions.
if the last number is 2 i.e. _ _ _ 2, the blanks can be filled by number 1, 6, 3 and they can be arranged in 3! ways.
In the same manner if the last number is 6 i.e. _ _ _ 6, the blanks can be filled by number 1, 2, 3 and they can be arranged in 3! ways.

Greatest number of four-digit extensions = 3!*2 = 12.
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Re: A math problem   [#permalink] 25 Jul 2011, 03:08
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