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m10q13

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Manager
Joined: 16 Feb 2010
Posts: 219

Kudos [?]: 355 [0], given: 16

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21 Nov 2010, 17:25
 How many odd three-digit integers greater than 800 are there such that all their digits are different?(C) 2008 GMAT Club - m10#13 * 40 * 56 * 72 * 81 * 104If the number begins with 8, there are $$5*8 = 40$$ possibilities (5 possibilities 1,3,5,7,9for the last digit and 8 possibilities for the middle digitwhy 8? it should be 9 (0,1,2,3,4,5,6,7,9)).If the number begins with 9, there are $$4*8 = 32$$ possibilities (4 possibilities for the last digit and 8 possibilities for the middle digitsame as above...it should be 9...).In all, there are $$40 + 32 = 72$$ numbers that satisfy the constraints.The correct answer is C.

thanks

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Manager
Joined: 25 Jun 2010
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21 Nov 2010, 18:27
You are missing that the last digit and middle digit can't be same.

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Manager
Joined: 16 Feb 2010
Posts: 219

Kudos [?]: 355 [0], given: 16

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21 Nov 2010, 18:39
anshumishra wrote:
You are missing that the last digit and middle digit can't be same.

Posted from my mobile device

gotta

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Re: m10q13   [#permalink] 21 Nov 2010, 18:39
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