Question 775339
I may be missing something but here is my calculation.
 
Zeros are a problem, because numbers sequences odf digits like 001 or 011 are not 3-digit numbers. So it is best to treat zero separately.
If zero is one of the digits, it could appear twice, necessarily at the end, in {{{highlight(red(9))}}} 3-digit numbers: 100, 200, ...800, and 900.
It could also appear once at the end in an additional {{{highlight(green(9))}}} 3-digit numbers: 110, 220, ...880, and 990.
It could also appear once in the middle in an additional {{{highlight(blue(9))}}} 3-digit numbers: 101, 202, ...808, and 909.
 
There are also {{{highlight(9)}}} 3-digit numbers made of the same repeating digit: 111, 222, ... 888, 999.
 
Besides the {{{red(9)+green(9)+blue(9)+9=highlight(36)}}} number we already listed, there are 3-digit numbers made of two non-zero digits, digit x appearing twice, and digit y appearing only once.
There are 9 ways to choose x, and for each of those there are 8 ways to choose y, for a total of {{{9*8=72}}} ordered pairs (x,y).
For each ordered pair, there are 3 positions where we can place y: at the beginning, in the middle, or at the end. That gives us
{{{72*3=216}}} 3-digit numbers made of two non-zero digits.
 
If I haven't missed any, and I have not counted any number more than once, that gives me
{{{216+36=highlight(252)}}} three-digit integers having at least one repeated digit.
 
With {{{system(a=2,b=5,c=2)}}} I get {{{a-b+c=2-5+2=highlight(-1)}}}