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vd
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P&C question 19 Jun 2008, 04:56
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Can u please tell me how to solve this

Many thanks
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greenoak
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P&C question 19 Jun 2008, 06:18
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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.
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FN
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P&C question 19 Jun 2008, 06:23
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greenoak..that was awesome..
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greenoak
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P&C question 19 Jun 2008, 06:39
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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. :(
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gmat2ndtime
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P&C question 19 Jun 2008, 06:42
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36 ^7 for me too....any other answers?
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jallenmorris
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P&C question 19 Jun 2008, 06:45
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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
36 ^7 for me too....any other answers?
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vd
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P&C question 19 Jun 2008, 23:20
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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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Hi there,
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