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# P&C question

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Senior Manager
Joined: 29 Aug 2005
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P&C question [#permalink]  19 Jun 2008, 03:56
Can u please tell me how to solve this

Many thanks
Attachment:
book.doc [28 KiB]

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Director
Joined: 12 Apr 2008
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Schools: Oxford
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Re: P&C question [#permalink]  19 Jun 2008, 05:18
Hmm, looks like quite easy.

For each digit, we have 26 letters + 10 numbers = 36 possibilities. There are 7 digits. So, total number of codes is 36*36*36*36*36*36*36 = 36^7.
Current Student
Joined: 28 Dec 2004
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Re: P&C question [#permalink]  19 Jun 2008, 05:23
greenoak..that was awesome..
Director
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Re: P&C question [#permalink]  19 Jun 2008, 05:39
I think that was an irony. But don't laugh at me.
I usually make mistakes in such questions In fact every time I see a probability/combinations problems, I'm dead.
Manager
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Re: P&C question [#permalink]  19 Jun 2008, 05:42
36 ^7 for me too....any other answers?
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Re: P&C question [#permalink]  19 Jun 2008, 05:45
I get E, $$36^7$$ too, but i'm kind of surprised the answer wasn't $$6^{14}$$ since 36 is a perfect square of 6.

I guess that depends on $$36^7$$ being the correct answer.

gmat2ndtime wrote:
36 ^7 for me too....any other answers?

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Senior Manager
Joined: 29 Aug 2005
Posts: 283
Followers: 1

Kudos [?]: 30 [0], given: 0

Re: P&C question [#permalink]  19 Jun 2008, 22:20
Guys the OA for this one is 36^7

any ideas on what basic material i can cover up so that i polish my P&C and Probability skills
Do the manhattan books help.

Many thanks
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The world is continuous, but the mind is discrete

Re: P&C question   [#permalink] 19 Jun 2008, 22:20
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