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December 19, 2014

December 19, 2014

Posted by **Kiki** on Friday, August 23, 2013 at 1:30am.

- Math -
**drwls**, Friday, August 23, 2013 at 1:59amAlready answered.

math - MathMate, Saturday, May 28, 2011 at 11:00pm

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

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