(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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$10$ men and $6$ women are to be seated in a row so that no two women sit together. The number of ways they can be seated is:

How many words can be formed using the letters of the word $INDEPENDENCE$,such that all vowels always come together?

How many distinct $9$-digit numbers can be formed by rearranging the digits of the number $223355888$ such that all odd digits occupy even positions?

In a class of $20$ students,in how many ways can the following prizes be awarded?If the first and second prizes are for Mathematics,the first and second prizes for Physics,the first prize for Chemistry,and the first prize for English are to be awarded.

Find the value of $n$ such that $\frac{^{n}P_{4}}{^{n-1}P_{4}} = \frac{5}{3}$,where $n > 4$.

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In a class of $20$ students,in how many ways can the following prizes be awarded?
If the first and second prizes are for Mathematics,the first and second prizes for Physics,the first prize for Chemistry,and the first prize for English are to be awarded.