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?
Question
This topic has expert replies
- rishimaharaj
- Senior | Next Rank: 100 Posts
- Posts: 90
- Joined: Mon May 02, 2011 11:18 am
- Location: Florida
- Thanked: 20 times
- Followed by:7 members
- GMAT Score:710
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
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