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

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

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New post 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


Please explain your answer.
thanks

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Re: PS: Interesting problem [#permalink]

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New post 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


Please explain your answer.
thanks

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Re: PS: Interesting problem [#permalink]

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New post 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


Please explain your answer.
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

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Re: PS: Interesting problem [#permalink]

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New post 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

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Re: PS: Interesting problem [#permalink]

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New post 05 Aug 2008, 12:29
thanks guys. OA is A

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Re: PS: Interesting problem [#permalink]

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New post 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


Please explain your answer.
thanks


7C2 = 7 options chose 2 places =
21.
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Re: PS: Interesting problem [#permalink]

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New post 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


Please explain your answer.
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.
:-D
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Re: PS: Interesting problem [#permalink]

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New post 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


Please explain your answer.
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.
:-D

mistake
2C7 is better i overlooked the conditions common between all. :roll:
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Re: PS: Interesting problem   [#permalink] 05 Aug 2008, 17:34
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Larry tried to type his new 7-digit phone number on a form,

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