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How many odd three-digit integers greater than 800 are there

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How many odd three-digit integers greater than 800 are there [#permalink] New post 31 Oct 2008, 07:49
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How many odd three-digit integers greater than 800 are there such that all their digits are different?

Please show your calculations. I will provide the OA.
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Re: three-digit integers [#permalink] New post 31 Oct 2008, 10:05
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That's a tough one. I get 68...
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Re: three-digit integers [#permalink] New post 31 Oct 2008, 10:48
72 for me

the Hundreds digit can be 8,9
Units can be 1,3,5,7,9
middle can be any of the 10

case 1 :

Hundreds digit = 8
Units can be any 5 digits
middle one can be any of the 8 digits apart from 8 & (any one of 1,3,5,7,9)

Total = 8*5=40

case 2 :

Hundreds digit = 9
Units can be any 4 digits (exclude 9)
middle one can be any of the 8 digits apart from 9 & (any one of 1,3,5,7)

Total = 8*4=32

Total no. of ways = 72
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Re: three-digit integers [#permalink] New post 31 Oct 2008, 13:05
From 800-899= Consider the tenth and unit position 1 (8)1 X even X odd= 1x4x5=20
Similarly =(8)1x odd x other odd=1x5x4=20
From 899-999= (9) 1x even x odd=1x5x4=20
Similarly=(9)1x oddx odd= 1x4x3=12 (because 9 is already used in the hundredth position.
So the total number of digits are=20+20+20+12=72
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Re: three-digit integers [#permalink] New post 01 Nov 2008, 06:56
subarao wrote:
From 800-899= Consider the tenth and unit position 1 (8)1 X even X odd= 1x4x5=20
Similarly =(8)1x odd x other odd=1x5x4=20

From 899-999= (9) 1x even x odd=1x5x4=20
Similarly=(9)1x oddx odd= 1x4x3=12 (because 9 is already used in the hundredth position.
So the total number of digits are=20+20+20+12=72


good one subarao :)
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Re: three-digit integers   [#permalink] 01 Nov 2008, 06:56
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