How many three-digit numbers are there such that all three digits are different and the first digit is not zero?
504
648
720
729
810
I know the numers to consider are form1 to 9. 9 Digits. Please explain
Combination
This topic has expert replies
-
- Legendary Member
- Posts: 869
- Joined: Wed Aug 26, 2009 3:49 pm
- Location: California
- Thanked: 13 times
- Followed by:3 members
-
- Master | Next Rank: 500 Posts
- Posts: 182
- Joined: Sun Aug 02, 2009 7:19 pm
- Thanked: 18 times
- GMAT Score:680
Since its a 3 digit numbe,obviously the 1st digit cannot be 0.
For the 1st digit : 1-9 - 9 ways
For the 2nd digit : we can pick from 0 and the remining 8 digits after the 1st is picked : again 9 ways
For the 3rd digit : 8 ways
Hence total number of 3 digit numbers : 9*9*8 = 648,choose (b)
For the 1st digit : 1-9 - 9 ways
For the 2nd digit : we can pick from 0 and the remining 8 digits after the 1st is picked : again 9 ways
For the 3rd digit : 8 ways
Hence total number of 3 digit numbers : 9*9*8 = 648,choose (b)
-
- Legendary Member
- Posts: 869
- Joined: Wed Aug 26, 2009 3:49 pm
- Location: California
- Thanked: 13 times
- Followed by:3 members
sanjana wrote:Since its a 3 digit numbe,obviously the 1st digit cannot be 0.
For the 1st digit : 1-9 - 9 ways
For the 2nd digit : we can pick from 0 and the remining 8 digits after the 1st is picked : again 9 ways
For the 3rd digit : 8 ways
Hence total number of 3 digit numbers : 9*9*8 = 648,choose (b)[/quote
I don't understand. After picking the first number, we have 8 left. How we have nine after picking the first 1 from original 9 digits?
First time you dont count 0. Second time you can count 0 (So actually 10 digits - 0 to 9). But, since digits cant repeat you have 9 again.heshamelaziry wrote: I don't understand. After picking the first number, we have 8 left. How we have nine after picking the first 1 from original 9 digits?
-
- Master | Next Rank: 500 Posts
- Posts: 399
- Joined: Wed Apr 15, 2009 3:48 am
- Location: india
- Thanked: 39 times
heshamelaziry wrote:WE HAVE IN TOTAL 10 DIGITS 0,1,2...9sanjana wrote:Since its a 3 digit numbe,obviously the 1st digit cannot be 0.
For the 1st digit : 1-9 - 9 ways
For the 2nd digit : we can pick from 0 and the remining 8 digits after the 1st is picked : again 9 ways
For the 3rd digit : 8 ways
Hence total number of 3 digit numbers : 9*9*8 = 648,choose (b)[/quote
I don't understand. After picking the first number, we have 8 left. How we have nine after picking the first 1 from original 9 digits?
FOR 1ST ONE WE CANT PICK 0 SO TOTAL WE HAVE 9 DIGITS SO 9 WAYS
FOR 2ND WE CAN INCLUDE SO AGAIN 9 DIGITS SO 9 WAYS
FOR 3RD ONLY 8 DIGITS R LEFT SO 8 WAYS
TOT 9*9*8=648
It does not matter how many times you get knocked down , but how many times you get up