If $P (n - 1, 3) : P (n, 4) = 1 : 9$ , find $n$ .

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Solution

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

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