(A) Given the ratio $^{n-1}P_{3} : ^{n}P_{4} = 1 : 9$. Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have: $\frac{^{n-1}P_{3}}{^{n}P_{4}} = \fr

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(A) Given the ratio $^{n-1}P_{3} : ^{n}P_{4} = 1 : 9$. Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have: $\frac{^{n-1}P_{3}}{^{n}P_{4}} = \frac{1}{9}$ $\Rightarrow \frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9}$ $\Rightarrow \frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = \frac{1}{9}$ $\Rightarrow \frac{(n-1)!}{n \times (n-1)!} = \frac{1}{9}$ $\Rightarrow \frac{1}{n} = \frac{1}{9}$ $\therefore n = 9$.

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

Medium

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

A
$9$
B
$10$
C
$8$
D
$11$

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Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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

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