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# PS-Comb

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Senior Manager
Joined: 05 Jun 2008
Posts: 289

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02 Dec 2008, 05:00
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(N/A)

Question Stats:

100% (00:02) correct 0% (00:00) wrong based on 3 sessions

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How many 5-digit positive integers exist where no two consecutive digits are the same?

A.) 9*9*8*7*6
B.) 9*9*8*8*8
C.) 9^5
D.) 9*8^4
E.) 10*9^4

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Current Student
Joined: 28 Dec 2004
Posts: 3280
Location: New York City
Schools: Wharton'11 HBS'12

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02 Dec 2008, 10:30
i get C as well

first integer is 9 ways, second integer 9 also, includes 0, so now the next we can continue the same 9^5
Senior Manager
Joined: 05 Jun 2008
Posts: 289

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03 Dec 2008, 05:48
scthakur wrote:
I think the answer should be A.

The first place has 9 possibilities, since 0 is not to be counted. All others have 9 each, since you cannot have the digit, which is same as the preceding one.
Hence 9^5
VP
Joined: 17 Jun 2008
Posts: 1474

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04 Dec 2008, 03:46
vivektripathi wrote:
scthakur wrote:
I think the answer should be A.

The first place has 9 possibilities, since 0 is not to be counted. All others have 9 each, since you cannot have the digit, which is same as the preceding one.
Hence 9^5

Oh, my silly analysis!

Thanks Vivek.
Intern
Joined: 01 Nov 2005
Posts: 43
Location: New jersey

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04 Dec 2008, 04:27
Whats the OA. i would think the answer would be since it is 5 digits .

first digit is going to be from 0..9
second digit is going to be one option less then 0..9
third .. (same as first two ) and so forth.

10*9*8*7*6

Curious to know the OA and explanation.
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Senior Manager
Joined: 30 Nov 2008
Posts: 479
Schools: Fuqua

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04 Dec 2008, 09:33
I guess the answer would be D - 9 * 8^4 And here is the explanation

Let _ _ _ _ _ be five digits positive integers (Means they can range from 1 to 9. I call them as first digit, second digit etc starting from Left side.

So the first digit can be picked from 9 positive integers = 9
Since no two consecutive digits should be same, second digit can be picked from 8 different numbers(eliminating the number picked from first digit = 8
Similary third, fourth and fifth digits can be picked from 8 different numbers each(eliminating the number picked from previous digit ).

So the answer is 9 * 8 * 8 * 8 * 8 = 9 * 8^4.
SVP
Joined: 29 Aug 2007
Posts: 2420

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04 Dec 2008, 10:07
mrsmarthi wrote:
I guess the answer would be D - 9 * 8^4 And here is the explanation

Let _ _ _ _ _ be five digits positive integers (Means they can range from 1 to 9. I call them as first digit, second digit etc starting from Left side.

So the first digit can be picked from 9 positive integers = 9
Since no two consecutive digits should be same, second digit can be picked from 8 different numbers(eliminating the number picked from first digit = 8
Similary third, fourth and fifth digits can be picked from 8 different numbers each(eliminating the number picked from previous digit ).

So the answer is 9 * 8 * 8 * 8 * 8 = 9 * 8^4.

First digit from 1 to 9 = 9 ways
Second digit from 0 to 9 but exclude one that is already in first digit = 9 ways
Third digit from 0 to 9 but exclude one that is already in second digit = 9 ways
Fourth digit is also from 0 to 9 but exclude one that is already in third digit = 9 ways
Fifth (final) digit is also from 0 to 9 but exclude one that is already in fourth digit = 9 ways

So the possibilities are 9x9x9x9x9 = 9^5 ways..

vivektripathi wrote:
How many 5-digit positive integers exist where no two consecutive digits are the same?

A.) 9*9*8*7*6
B.) 9*9*8*8*8
C.) 9^5
D.) 9*8^4
E.) 10*9^4

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Senior Manager
Joined: 30 Nov 2008
Posts: 479
Schools: Fuqua

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04 Dec 2008, 13:09
I have given my guess based on the fact that it is 5 digit positive integers. So my understanding is all the 5 digits can be positive integers in other words 1 to 9. O is neither positive or negative.

SVP
Joined: 29 Aug 2007
Posts: 2420

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04 Dec 2008, 14:49
mrsmarthi wrote:
I have given my guess based on the fact that it is 5 digit positive integers. So my understanding is all the 5 digits can be positive integers in other words 1 to 9. O is neither positive or negative.

You are talking about a single digit where as the question is about 5 digit integer.
Suppose if 10,101 is a 5 digit +ve integer, -10,101 is a 5 digit -ve integer.

This is what meant by the +ve and -ve integer in the question.

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Re: PS-Comb &nbs [#permalink] 04 Dec 2008, 14:49
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