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Manager
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m10q13 [#permalink] New post 21 Nov 2010, 17:25
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wrote:
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
* 104

If 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
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Re: m10q13 [#permalink] New post 21 Nov 2010, 18:27
You are missing that the last digit and middle digit can't be same.

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Re: m10q13 [#permalink] New post 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 Image


gotta ;-)
Re: m10q13   [#permalink] 21 Nov 2010, 18:39
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