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

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Solution

$\begin{matrix} \frac{^{n - 1} P_{3}}{^{n} P_{4}} = \frac{1}{9} \Rightarrow \frac{[\frac{(n - 1)!}{3!}]}{[\frac{n!}{4!}]} = \frac{1}{9} \Rightarrow \frac{(n - 1)!}{3!} \times \frac{4 \times 3!}{n (n - 1)!} \Rightarrow \frac{4}{n} = \frac{1}{9} n = 36 \end{matrix}$

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Similar questions

^{n - 1} P_{3} :^{n} P_{4} = 1.4

n

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

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

5. {^{n} P}_{3} = 4. {^{n + 1} P}_{3}

n

p (x) = a x^{2} + b x

q (x) = l x^{2} + m x + n

p (1) = q (1); p (2) - q (2) = 1

p (3) = q (3) = 4

p (4) - q (4)

m / n

m - (8 * n)

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