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Q. How many 5-digit positive integers exist where no two

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Q. How many 5-digit positive integers exist where no two [#permalink] New post 17 Feb 2007, 20:04
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Q. 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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Re: PS consecutive digits [#permalink] New post 17 Feb 2007, 20:34
sgoll wrote:
Q. 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


1st digit can be 1 to 9 = 9 ways
2nd digit can be 0 to 8 = 9 ways (we consumed 1 digit already in the 1st didgit)
3rd digit can be 0 to 7 = 8 ways
4th digit can be 0 to 6 = 7 ways
5th digit can be 0 to 5 = 6 ways

Total # of numbers = 9*9*8*7*6 (A)
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Re: PS consecutive digits [#permalink] New post 17 Feb 2007, 20:35
sgoll wrote:
Q. 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


let the number be abcde

a can be 1-9 so 9 choices
b can be 0-9 but not same as a so 9 choices
c can be 0-9 but not same as b so 9 choices
so on for d,e

so ans is 9^5 - C
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Re: PS consecutive digits [#permalink] New post 17 Feb 2007, 20:37
prude_sb wrote:
sgoll wrote:
Q. 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


let the number be abcde

a can be 1-9 so 9 choices
b can be 0-9 but not same as a so 9 choices
c can be 0-9 but not same as b so 9 choices
so on for d,e

so ans is 9^5 - C


prude: Can you explain the thing in bold please? Thanks !
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Re: PS consecutive digits [#permalink] New post 17 Feb 2007, 20:48
trivikram wrote:
prude_sb wrote:
sgoll wrote:
Q. 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


let the number be abcde

a can be 1-9 so 9 choices
b can be 0-9 but not same as a so 9 choices
c can be 0-9 but not same as b so 9 choices
so on for d,e

so ans is 9^5 - C


prude: Can you explain the thing in bold please? Thanks !


a can be any number from 1 to 9
b can be anything other than 'a' so it can be any number from 0-9 other than a
c can be any number other than 'b' it can also be same as 'a'
similarly for d,e

32323 would be acceptable since we only need the consecutive integers to be different

choice A gives the number of cases where ALL digits are unique ...
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 [#permalink] New post 17 Feb 2007, 21:25
Yes OA is C. Thanks for the detailed explanation prude_sb
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 [#permalink] New post 18 Feb 2007, 05:48
Got it...Again read the question wrong....
  [#permalink] 18 Feb 2007, 05:48
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Q. How many 5-digit positive integers exist where no two

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