How many 3-digit numbers can be formed under the following conditions:
a)leading digit cannot be zero
There are 999 positive integers from 1 through 999
There are 99 positive integers from 1 to 99, which we cannot count.
Therefore, the number of 3 digit numbers is 999-99 or 900
b)leading digit cannot be zero and no repetition of digits is allowed
There are 9 ways to choose for the first digit, since we cannot choose 0.
For each of those 9 ways to choose the 1st digit, we can choose
the second digit also 9 ways. That's because even though we cannot
choose the same digit we chose for the 1st digit, we CAN choose 0,
whereas we could not choose 0 for the 1st digit.
So that's 9×9 or 81 ways to choose the first two digits.
For each of those 9×9 or 81 ways to choose the first two digits, there
are 8 ways to choose the last digit.
Answer = 9×9×8 = 81×8 = 648 ways
c)the number is at least 400
There are 999 positive integers with 3 or fewer digits,
there are 399 which are not at least 400, and all the rest have 3 digits.
Therefore the answer = 999-399 = 600
Edwin