I to be considering strings of 4 decimal digits that contain the very same digit twice. V this, I have the possibilities that $xxyz,xyxz,yxxz,yxzx,yzxx,xyzx$ wherein $x$ is the exact same digit, and also $y, z$ room randomly various decimal number taken. For this reason $10 cdot 1 cdot 9 cdot 8 = 720$ and also $6$ possibilities, then $720 cdot 6 = 4320$ ways. Now consider digits that have actually pattern $xxyy, yyxx,xyxy,yxxy,yxyx,xyyx$, here $10 cdot 1 cdot 9 cdot 9=810 cdot 6= 4860$ ways. Then total combinations space $10^4 =10000$ climate $10000-4320-4860=820$ means .

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

Where it pertains to strings that have $3$ unique digits your calculation through outcome $$inom42cdot10cdot9cdot8=4320$$ is correct.

Where it comes to strings that have $2$ unique digits that both appear twice the outcome have to be: $$frac12inom42cdot10cdot9=270$$You (maybe accidently) have $2$ determinants $9$ rather of $1$ and also did not repair dual counting.

You could also reason the there space not $6$ however $3$ fads there ($xxyy$,$xyxy$ and also $xyyx$) and this v $10$ selections for $x$ and also $9$ remaining choices for $y$.


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How plenty of ways room there to select strings of length $3$ native $S=1,2,3,4$ because that $3$ people if each each digit from $S$ has to show up at least once?

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