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: \$\$frac12inom42cdot10cdot9=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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