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# Larry tried to type his new 7-digit phone number on a form,

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VP
Joined: 21 Jul 2006
Posts: 1485
Larry tried to type his new 7-digit phone number on a form, [#permalink]

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05 Aug 2008, 11:46
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Larry tried to type his new 7-digit phone number on a form, but what appeared on the form was 39269, since the '4' key on his computer no longer works. His secretary has decided to make a list of all of the numbers that could be Larry's new number. How many numbers will there be on the list?

a) 21
b) 24
c) 25
d) 30
e) 36

thanks

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SVP
Joined: 30 Apr 2008
Posts: 1850
Location: Oklahoma City
Schools: Hard Knocks

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05 Aug 2008, 11:56
i think A.

Essentially we have 7 options. We need to find out how many different variations there are for 2 of those numbers. That seems like it would be $$C_7^2 = 21$$.

tarek99 wrote:
Larry tried to type his new 7-digit phone number on a form, but what appeared on the form was 39269, since the '4' key on his computer no longer works. His secretary has decided to make a list of all of the numbers that could be Larry's new number. How many numbers will there be on the list?

a) 21
b) 24
c) 25
d) 30
e) 36

thanks

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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

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Intern
Joined: 04 Jun 2008
Posts: 16
Concentration: Marketing, Strategy
GMAT 1: 690 Q48 V35
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WE: Engineering (Non-Profit and Government)

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05 Aug 2008, 12:03
tarek99 wrote:
Larry tried to type his new 7-digit phone number on a form, but what appeared on the form was 39269, since the '4' key on his computer no longer works. His secretary has decided to make a list of all of the numbers that could be Larry's new number. How many numbers will there be on the list?

a) 21
b) 24
c) 25
d) 30
e) 36

thanks

7 options and 2 different instances of 4:

C(7,2) = 21

Longer way:

44.....
4.4....
4..4...
4...4..
4....4.
4.....4
.44....
.4.4...
.4..4..
.4...4.
.4....4
..44...
..4.4..
..4..4.
..4...4
...44..
...4.4.
...4..4
....44.
....4.4
.....44

=21

A
Senior Manager
Joined: 23 May 2006
Posts: 309

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05 Aug 2008, 12:04
Larry's typed the seven digits, however since only (39269) five digits appeared the rest fo the two digits must be two 4's.
Now I think if we can rephrase the question by saying - how many ways can we arrange two numbres in seven slots.

7c2 = 21.

Not sure if this is the right answer.

IMO A
VP
Joined: 21 Jul 2006
Posts: 1485

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05 Aug 2008, 12:29
thanks guys. OA is A
SVP
Joined: 07 Nov 2007
Posts: 1756
Location: New York

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05 Aug 2008, 12:37
tarek99 wrote:
Larry tried to type his new 7-digit phone number on a form, but what appeared on the form was 39269, since the '4' key on his computer no longer works. His secretary has decided to make a list of all of the numbers that could be Larry's new number. How many numbers will there be on the list?

a) 21
b) 24
c) 25
d) 30
e) 36

thanks

7C2 = 7 options chose 2 places =
21.
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VP
Joined: 17 Jun 2008
Posts: 1322

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05 Aug 2008, 17:31
tarek99 wrote:
Larry tried to type his new 7-digit phone number on a form, but what appeared on the form was 39269, since the '4' key on his computer no longer works. His secretary has decided to make a list of all of the numbers that could be Larry's new number. How many numbers will there be on the list?

a) 21
b) 24
c) 25
d) 30
e) 36

thanks

Number of ways to put a single number in 5 digits(different from the number) is 6.Now consider this 6 digit number hence formed and here to append one number same as previous is 6 again excluding the possibility off two same numbers together two times.
hence 6*6=36 combinations are possible.

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VP
Joined: 17 Jun 2008
Posts: 1322

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05 Aug 2008, 17:34
spriya wrote:
tarek99 wrote:
Larry tried to type his new 7-digit phone number on a form, but what appeared on the form was 39269, since the '4' key on his computer no longer works. His secretary has decided to make a list of all of the numbers that could be Larry's new number. How many numbers will there be on the list?

a) 21
b) 24
c) 25
d) 30
e) 36

thanks

Number of ways to put a single number in 5 digits(different from the number) is 6.Now consider this 6 digit number hence formed and here to append one number same as previous is 6 again excluding the possibility off two same numbers together two times.
hence 6*6=36 combinations are possible.

mistake
2C7 is better i overlooked the conditions common between all.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________

cheers
Its Now Or Never

Re: PS: Interesting problem   [#permalink] 05 Aug 2008, 17:34
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