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If $\mathrm{n}, \mathrm{r}$ are two positive integers such that $1 \leq \mathrm{r} < \mathrm{n}$, then ${ }^n P_{r+1}+r^2 n-1 P_{r-1}+(r+1)^{n-1} P_r+r^{n-1} P_{r-1}=$
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The correct answer is:
${ }^{n+1} P_{r+1}$
${ }^n P_{r+1}+r^2 \cdot{ }^{n-1} P_{r-1}+(r+1){ }^{n-1} P_r+r \cdot{ }^{n-1} P_{r-1}$
$\begin{aligned} & ={ }^n P_{r+1}+(r+1){ }^{n-1} \cdot P_r+r^2 \cdot{ }^{n-1} P_{r-1}+r \cdot{ }^{n-1} P_{r-1} \\ & =\frac{n !}{(n-r-1) !}+(r+1) \cdot \frac{(n-1) !}{(n-r-1) !}+r^2 \cdot \frac{(n-1) !}{(n-r) !}+r \cdot \frac{(n-1) !}{(n-r) !} \\ & =\frac{n !(n-r)+(r+1)(n-1) !(n-r)+r^2(n-1) !+r(n-1) !}{(n-r) !} \\ & =\frac{(n-1) !\left(n(n-r)+(r+1)(n-r)+r^2+r\right)}{(n-r) !} \\ & =\frac{(n-1) !\left(n^2-n r+n r-r^2+n-r+r^2+r\right)}{(n-r) !} \\ & =\frac{(n-1) !\left(n^2+n\right)}{(n-r) !}=\frac{n(n+1)(n-1) !}{(n-r) !} \\ & =\frac{(n+1) !}{(n-r) !}={ }^{n+1} P_{r+1} \cdot\end{aligned}$
$\begin{aligned} & ={ }^n P_{r+1}+(r+1){ }^{n-1} \cdot P_r+r^2 \cdot{ }^{n-1} P_{r-1}+r \cdot{ }^{n-1} P_{r-1} \\ & =\frac{n !}{(n-r-1) !}+(r+1) \cdot \frac{(n-1) !}{(n-r-1) !}+r^2 \cdot \frac{(n-1) !}{(n-r) !}+r \cdot \frac{(n-1) !}{(n-r) !} \\ & =\frac{n !(n-r)+(r+1)(n-1) !(n-r)+r^2(n-1) !+r(n-1) !}{(n-r) !} \\ & =\frac{(n-1) !\left(n(n-r)+(r+1)(n-r)+r^2+r\right)}{(n-r) !} \\ & =\frac{(n-1) !\left(n^2-n r+n r-r^2+n-r+r^2+r\right)}{(n-r) !} \\ & =\frac{(n-1) !\left(n^2+n\right)}{(n-r) !}=\frac{n(n+1)(n-1) !}{(n-r) !} \\ & =\frac{(n+1) !}{(n-r) !}={ }^{n+1} P_{r+1} \cdot\end{aligned}$
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