If P(n, 4) = 12 . P(n, 2), find n.

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We have,
P(n, 4) = 12 . P(n, 2).
$\Rightarrow \frac{\text{n!}}{(\text{n-4})!}=12\times\frac{\text{n!}}{(\text{n-2)}! }\ \Big[\because^\text{n}\text{p}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$
$\Rightarrow \frac{1}{(\text{n-4})!}=\frac{12}{(\text{n-2)}!}$
$\Rightarrow \frac{1}{(\text{n}-4)}!= \frac{12}{(\text{n}-2)(\text{n}-2 -1)(\text{n}-2-2)!}$
$\Rightarrow \frac{1}{(\text{n}-4)!}= \frac{12}{(\text{n}-2)(\text{n}-3)\text{n}-4)!}$
$\Rightarrow \frac{(\text{n}-2)(\text{n}-3)(\text{n}-4)!}{(\text{n}-4)!}= 12$
$\Rightarrow (\text{n}-2)(\text{n}-3) =12$
$\Rightarrow \text{n}^2-3\text{n}-2\text{n}-12=0$
$\Rightarrow \text{n}^2+5\text{n}-6=0$
$\Rightarrow \text{n}^2-6\text{n}+1\text{n}-6=0$
$\Rightarrow \text{n}(\text{n}-6)+1(\text{n}-6)= 0$
$\Rightarrow (\text{n}-6)(\text{n}+1)= 0$
$\Rightarrow \text{n}-6=0$
$\Rightarrow \text{n}=6 \ [\because \text{n}\neq-1]$
Hence, $\text{n}=6$

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