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gilm: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Sun Aug 07, 2011 2:06 am; Thanked: 1 times

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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Source: Beat The GMAT — Problem Solving | Tue Aug 09, 2011 7:22 am

rishimaharaj: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Mon May 02, 2011 11:18 amFollowed by: 7 members; Location: Florida; Thanked: 20 times; GMAT Score:710

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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