Question:
If $n+5 P_{n+1}=\frac{11(n-1)}{2} n+3 P_{n}$, find $n$.
Solution:
${ }^{n+5} \mathrm{P}_{n+1}=\frac{11(n-1)}{2} n+3 \mathrm{P}_{n}$
$\Rightarrow \frac{(n+5) !}{(n+5-n-1) !}=\frac{11(n-1)}{2} \times \frac{(n+3) !}{(n+3-n) !}$
$\Rightarrow \frac{(n+5) !}{4 !}=\frac{11(n-1)}{2} \times \frac{(n+3) !}{3 !}$
$\Rightarrow \frac{(n+5) !}{(n+3) !}=\frac{11(n-1)}{2} \times \frac{4 !}{3 !}$
$\Rightarrow \frac{(n+5)(n+4)(n+3) !}{(n+3) !}=\frac{11(n-1)}{2} \times \frac{4 \times 3 !}{3 !}$
$\Rightarrow(n+5)(n+4)=22(n-1)$
$\Rightarrow n^{2}+9 n+20=22 n-22$
$\Rightarrow n^{2}-13 n+42=0$
$\Rightarrow n=7,6$
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