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Junior | Next Rank: 30 Posts
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by gilm » Tue Aug 09, 2011 7:22 am
How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other? a) 648. b) 504. c) 576. d) 810. e) 672.

I figure this to be 9*10*9=810 because it says "other digits" which I understand as the two digits other than the first. However, the answer is 9*9*8=648

Can anyone explain?
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Senior | Next Rank: 100 Posts
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by rishimaharaj » Mon Apr 30, 2012 7:56 am
First digit != 0 := {1,2,3,4,5,6,7,8,9} which is 9 numbers.
Second digit != first digit := 10 total digits minus first digit = 9 possible numbers.
Third digit != first two digits := 10 total digits minus 2 already used = 8 possible numbers.

This results in 9 * 9 * 8 = 648.

Hope this helps!
--Rishi
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