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If ${ }^n P_r={ }^{(n-1)} P_r+x^{(n-1)} P_{(r+1)}, \forall n, r \in N$ and $r \leq n$, then $x=$
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$r$
$\begin{aligned}{ }^n P_r- & { }^{n-1} P_r=\frac{n !}{(n-r) !}-\frac{(n-1) !}{(n-r-1) !} \\ & =\frac{(n-1) !}{(n-r-1) !}\left(\frac{n}{n-r}-1\right)=\frac{(n-1) !}{(n-r) !} \cdot r \\ & =\frac{(n-1) !}{((n-1)-(r-1) !} \cdot r=r \cdot{ }^{n-1} P_{r-1} \\ \therefore \quad{ }^n P_r & ={ }^{n-1} P_r+r \cdot{ }^{n-1} P_{r-1} \\ \therefore \quad x & =r\end{aligned}$
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