How many $n$-digit decimal sequences (using the digits $0 = 9$) are there in which the digits $1$, $2$ and $3$ all appear?

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I feel the answer should be $nC3 \times 3! \times 10^{n-3}$.

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1 Answer

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The number of $n$ digit sequences is $10^n$. The number missing $1$ is $9^n$, so we would subtract $3\cdot 9^n$ to remove the ones missing $1, 2, \text { or }3$, but now we have removed the ones missing $1$ and $2$ (or any other pair) twice so we have to add back in $3 \cdot 8^n$. Now the ones missing all of $1,2,3$ have been removed three times and added back three times, so we subtract them once more. Finally we have $10^n-3\cdot 9^n + 3 \cdot 8^n-7^n$. This is known as the inclusion-exclusion principle

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