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

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Solution

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

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

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

$\Rightarrow n = 9$

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Similar questions

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

{(n - 1)}_{p_{3}} : n_{p_{4}} = 1 : 4

n

(i) If $^{n} P_{4} :^{n} P_{5} = 1 : 2$ find n.

(ii) If $^{n - 1} P_{3} :^{n + 1} = 5 : 12,$ find n.

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

^{n} P_{4} = 5 (^{n} P_{3})

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