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GmatPrep [#permalink]

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New post 22 Jan 2009, 07:24
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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?
21
24
25
30
36
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New post 22 Jan 2009, 09:57
A

1. 39269 contains 5 digits. Therefore, there are 2 '4' digits.
2. 2 '4' digits we can place in 7 possible positions by 7C2 different positions (we use 'C' instead of 'P' as a number like xx4x4x is insensitive to permutation of '4's)

7C2=7*6/2=21
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New post 23 Jan 2009, 13:07
I did it with a different approach and reached another answer:

1. we need to add two 4's.
2. we have 6 places to add them:

_3_9_2_6_9_

so we have 6 places to put the first 4, and again 6 places for the second 4.
so 6x6=36.
What am I missing?
What is the OA?

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New post 23 Jan 2009, 15:07
I think you forget to subtract double counting. We can divide your 36 ways to get a number by two groups.

1) First group, a pair group contains two '4' together like xx44xxx. We have 6 places and 6 ways to get such kind of a number. Moreover, this group doesn't contain double counting.

2) Second group, a separated group contains two '4' separated by other digit(s) like x4xx4xx. We have 36-6=30 ways to get such kind of a number. But in your reasoning you get the same number twice: x(first placed '4')xx(second placed '4')xx and x(second placed '4')xx(first placed '4')xx represent two ways but only one unique number.

Therefore, we should count numbers as 6+30/2=6+15=21
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New post 24 Jan 2009, 03:10
Got it.
Thanks.

Could you explain about the 7 places at your first method?

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New post 24 Jan 2009, 03:17
"Larry tried to type his new 7-digit phone number on a form ...." - 7 digits - 7 places for two '4'. Other 5 places are filled by 39269
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New post 24 Jan 2009, 03:48
How do you solve such questions under pressure during the test?! :)
It took me more than 2 min to understand what the question was trying to say.

So: there are 6 ways to insert the first 4 and 7 ways to insert the second one. You need to divide the result by 2 as half of the time you will have the same number you already had. (6x7)/2= 21

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New post 31 Jan 2009, 21:40
walker wrote:
A

1. 39269 contains 5 digits. Therefore, there are 2 '4' digits.
2. 2 '4' digits we can place in 7 possible positions by 7C2 different positions (we use 'C' instead of 'P' as a number like xx4x4x is insensitive to permutation of '4's)

7C2=7*6/2=21


Many thanks, OA is A
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Re: GmatPrep   [#permalink] 31 Jan 2009, 21:40
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