Show that:
$^{n} P_{n} = 2^{n} P_{n - 2}$

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Solution

$\begin{matrix} ^{n} P_{n} = \frac{n!}{(n - r)!} H e n c e^{n} P_{n} = \frac{n!}{(n - r)!} = \frac{n!}{0!} = n! A l s o,^{n} P_{n} = \frac{n!}{(n - (n - 2))!} = \frac{n!}{(n - n + 2)!} = \frac{n!}{2!} = \frac{n!}{2} S o, 2 \times (\frac{n!}{2}) = n! =^{n} P_{n} H e n c e^{n} P_{n} = 2 \times \frac{n P}{(n - 2)} \end{matrix}$

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