Combination

[heshamelaziry]

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

[sanjana]

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)

[heshamelaziry]

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?

[enniguy]

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.

[xcusemeplz2009]

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?

WE HAVE IN TOTAL 10 DIGITS 0,1,2...9
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