(A) Given the equation $\frac{^{n}P_{4}}{^{n-1}P_{4}} = \frac{5}{3}$.Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:$\frac{\frac{n!}{(n-4)!

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(A) Given the equation $\frac{^{n}P_{4}}{^{n-1}P_{4}} = \frac{5}{3}$.Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:$\frac{\frac{n!}{(n-4)!}}{\frac{(n-1)!}{(n-5)!}} = \frac{5}{3}$.This simplifies to $\frac{n \times (n-1)!}{(n-4) \times (n-5)!} \times \frac{(n-5)!}{(n-1)!} = \frac{5}{3}$.$\frac{n}{n-4} = \frac{5}{3}$.Cross-multiplying gives $3n = 5(n-4)$.$3n = 5n - 20$.$2n = 20$.$n = 10$.

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

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

A
$10$
B
$12$
C
$8$
D
$15$

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

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Find the number of arrangements of the letters of the word $INDEPENDENCE$. In how many of these arrangements do the words start with $P$?

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