Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If all of the telephone extensions in a certain company must [#permalink]
22 Jan 2005, 13:03

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

100% (03:48) correct
0% (00:00) wrong based on 3 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?

Re: If all of the telephone extensions in a certain company must [#permalink]
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.

Re: PS telephones numbers [#permalink]
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

Re: PS telephones numbers [#permalink]
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.

[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

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

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!

Re: telephone extensions [#permalink]
09 Jun 2009, 15:04

1

This post received KUDOS

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.

Re: telephone extensions [#permalink]
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

Re: A math problem [#permalink]
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.

Re: A math problem [#permalink]
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.

Re: A math problem [#permalink]
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. _________________

Cheers, Varun

If you like my post, give me KUDOS!!

gmatclubot

Re: A math problem
[#permalink]
25 Jul 2011, 03:08