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M25 # 20

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M25 # 20 [#permalink] New post 10 Aug 2009, 14:53
Can someone give me a hand with understanding the formula behind this better?

I can't seem to wrap my head around it?

Thanks



How many odd three-digit integers greater than 800 are there such that all their digits are different?

(C) 2008 GMAT Club - m25#20

40
56
72
81
104
If the number begins with 8, there are possibilities (5 possibilities for the last digit and 8 possibilities for the middle digit). If the number begins with 9, there are possibilities (4 possibilities for the last digit and 8 possibilities for the middle digit). In all, there are numbers that satisfy the constraints.

The correct answer is C.
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Re: M25 # 20 [#permalink] New post 10 Aug 2009, 15:30
Number of odd digit numbers starting with 8.
Hundreds place is filled with 8.
For a number to be odd, units place should be filled with either 1, 3, 5, 7, 9.
So there are 5 possibilities.
Tens place can be filled with either 0,1,2,3,4,5,6,7,9 .
But we have to eliminate the digits that are used in units position.
So number of odd numbers with unique digits starting with 8 = 1*8*5 = 40 ways.

For numbers starting with 9, unit digit can be either or 1,3,5,7
So number of odd numbers with unique digits starting with 9 = 1*8*4 = 32 ways.

So total number of possible combinations = 40 + 32 = 72 ways
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Re: M25 # 20 [#permalink] New post 10 Aug 2009, 17:03
Expert's post
Already discussed: m25-76554.html

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Re: M25 # 20   [#permalink] 10 Aug 2009, 17:03
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