Count of number of $3$-digit strings with at least two of $0,1,2$. 
Find the count of $3$-digit strings with at least two of $0,1,2$.

Well I am getting $432$ as the answer, which I think is way too much.
For starters, I considered that one number of $\{0,1,2\}$ could take any place so $3 \times 3$.  Then any other number (previous number included) would take $3 \cdot 2$ so total rn is $3 \times 3 \times 3 \times 2$, now the last place can be filled by any of the $9$ digit so $\times 9$.
After reducing the repeated numbers ($9 \times 3 \times 2$), which according to me will all be palindromes with these numbers, I got my result.
Looks like I need help here.  If I did anything wrong, I'd be happy to hear your comments.
 A: There are $1000$ possible strings with no constraints whatsoever.  Of those, $7^3=343$ use none of the digits $\{0, 1, 2 \}$.
To count the strings that have an entry from $\{0, 1, 2 \}$ in exactly one position, note that we have $3$ choices of that position, $3$ choices from among $\{ 0, 1, 2 \}$ to fill that position, and $7^2=49$ choices to fill the remaining positions, for a total of $9 \cdot 49=441$ additional "bad" strings.
Thus, there are $1000-(343+441)=216$ "good" strings.
Edit:  As noted by fleablood in an answer and by JMoravitz in a comment, you can use the same method of analysis to count the "good" strings directly.  There are $3^3=27$ strings that fill all three positions from the set $\{0, 1, 2\}$.
To count the strings that have an entry from $\{4, 5, 6, 7, 8, 9, 0 \}$ in exactly one position (and therefore entries from $\{ 1, 2, 3 \}$ in exactly two positions), note that we have $3$ choices of that position, $7$ choices from among $\{ 4, 5, 6, 7, 8, 9, 0 \}$ to fill that position, and $3^2=9$ choices to fill the remaining positions for a total of $3 \cdot 7 \cdot 9 =189$ additional "good" strings.
Thus, there are $27+189 = 216$ good strings.
A: So going over your method.

i considered that one number of {0,1,2} could take any place so 3×3

If I understand you correctly your nine options are $0xx, x0x, xx0; 1xx,x1x,xx1; 2xx,x2x,xx2$.  Correct.  You are choosing the value of one of the numbers and choosing its position.

Then any other number (previous number included) would take 3×2 so total rn is 3×3×3×2

Now you are double counting.  If we mark the first chosen number in $\color{green}{\text{green}}$ and the second in $\color{red}{\text{red}}$.  Choosing first $1$ for position $3$ and $2$ for position $1$; that is:  $\color{red}2x\color{green}1$.  But if you chose first the $2$ for position $1$ and then $1$ for positon $3$ to get $\color{green}2x\color{red}1$ you are counting these as different choices.
As the order of the choices doesn't matter you must divide by $2$ to have $\frac{3\times 3\times 3\times 2}2=3\times 3\times 3$.

now the last place can be filled by any of the 9 digit so ×9

But now you are still over counting.  If third digit is $0,1,2$ then .... well let's put the first two digits in $\color{blue}{\text{blue}}$ and the third in $\color{purple}{\text{purple}}$ we have $\color{blue}{32}\color{purple}3$ and $\color{blue}3\color{purple}2\color{blue}3$ and $\color{purple}3\color{blue}{23}$ counting as three different numbers.
To handle this we mus consider for the $3$ cases were the third digit is $0,1$ or $2$ we must divide by the $3$ places the third digit can be.  So $(3\times3\times 3)\times \frac 33 = 3\times 3\times 3$.
Then for the $7$ cases where the third digit is not $0,1,2$ you aren't over counting so we have $3\times 3\times 3\times 7$.
Adding those together we have $3\times 3\times 3 + 3\times 3\times 3\times 3 = 216$.
.....
Although it might be better to make a simpler model.
IMO I'd do...
There are two cases.  Two of the three digits are $0,1,2$ and the third isn't.  Or all three of the three digits are $0,1,2$.
If all three of the digits are $0,1,2$ then for each position there are $3$ choices so the number ways is $choice^{positions} = 3^3 = 27$ (or $3\times 3\times 3$ if you prefer.
If one of the digits is not $0,1,$ or $2$ then there are $3$ choices for that position.  And there are $7$ choices for it's value. $(3\times 7)$.  Then for the remaining $2$ position there are $3$ choices for each value so $(3\times 7)\times 3^2$.
So the total number is $(3\times 3\times 3) + (3\times 7)\times(3\times 3) = 216$.
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Edit:  Oh.  Robert Shore's method of elimination is a good model too.  Maybe even easier.  Number of at least two = All numbers - Those with fewer than two = All number - (Those with none + those with exactly one).  Can't argue with that and those are relatively easy values to calculate.
I do find it interesting when numbers combine in strange ways to get same result.
$3^3 + 3\times 7\times 3^2 = 10^3 -(7^3 + 3\times 3\times 7^2)$.
Who'd have thunk it.
