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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?
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
