How many strings of four decimal digits do not contain the same digit twice?

Your title is ambiguous (see the comment of Mohammad) and I preassume that it concerns the number of strings that have the property that no digit appears exactly $2$ times.

Where it concerns strings that have $3$ distinct digits your calculation with outcome $$\binom42\cdot10\cdot9\cdot8=4320$$ is correct.

Where it concerns strings that have $2$ distinct digits that both appear twice the outcome should be: $$\frac12\binom42\cdot10\cdot9=270$$ You (maybe accidently) have $2$ factors $9$ instead of $1$ and did not repair double counting.

You could also reason that there are not $6$ but $3$ patterns there ($xxyy$,$xyxy$ and $xyyx$) and this with $10$ choices for $x$ and $9$ remaining choices for $y$.