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Evaluate ${}_nP_1$? I believe the answer is $n$. Am I correct?

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Well, do you know the formula for $_n P_r$? $$_nP_r=\frac{n!}{(n-r)!}$$ Just plug in the values of $n$ and $r$ into this formula. In this case your $n$ still equals $n$, and $r$ equals $1$. Therefore: $$_nP_1=\frac{n!}{(n-1)!}$$ $$=\frac{n(n-1)(n-2)\dots 3\cdot 2\cdot 1}{(n-1)(n-2)\dots 3\cdot 2\cdot 1}$$ $$=n$$ $$\color{green}{\boxed{_nP_1=n}}$$ Hope I helped

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