$^{n} p_{4} :^{n - 1} p_{3} = 9 : 1$ .Find n

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Solution

We know that, $^{n} P_{r} = \frac{n!}{(n - r)!}$

Given, $^{n} P_{4} :^{n - 1} P_{3} = 9 : 1$
$\Rightarrow \frac{^{n} P_{4}}{^{n - 1} P_{3}} = \frac{9}{1}$

$\Rightarrow \frac{\frac{n!}{(n - 4)!}}{\frac{(n - 1)!}{(n - 1 - 3)!}} = 9$
$\Rightarrow \frac{\frac{n (n - 1)!}{(n - 4)!}}{\frac{(n - 1)!}{(n - 4)!}} = 9$

$\Rightarrow n = 9$

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