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

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Solution

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

$∴ \frac{(n - 1)!}{(n - 4)!} : \frac{n!}{(n - 4)!} = 1 : 9$

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

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

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

$\Rightarrow \frac{1}{n} = \frac{1}{9} \Rightarrow n = 9$

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

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

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(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} = 18^{n - 1} P_{2}

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

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