posted by alex .
There are 900 three-digit integers. The number of three-digit integers having at least one repeated digit is also a three-digit number. If that number is represented by abc where each letter is a digit, compute
(I came up with 252 integers so when I did the computation it didn't make sense)
Here we have 10 digits, of which the first one cannot be zero.
If the three digits are distinct, there are 9 choices for the first digit, still 9 (including zero) for the second, and 8 for the third.
So 9*9*8=648 numbers.
There are therefore 900-648=252 numbers which have at least one repeated digit.
So a=2, b=5, c=2, and a-b+c=2-5+2=-1