(D) Given the equation: $^n{P_5} = 9 \times {^{n - 1}}{P_4}$Using the formula $^n{P_r} = \frac{n!}{(n-r)!}$,we have:$\frac{n!}{(n-5)!} = 9 \times \fra

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(D) Given the equation: $^n{P_5} = 9 \times {^{n - 1}}{P_4}$Using the formula $^n{P_r} = \frac{n!}{(n-r)!}$,we have:$\frac{n!}{(n-5)!} = 9 \times \frac{(n-1)!}{(n-1-4)!}$$\frac{n \times (n-1)!}{(n-5)!} = 9 \times \frac{(n-1)!}{(n-5)!}$Since $(n-1)!$ and $(n-5)!$ are non-zero,we can cancel them from both sides:$n = 9$

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

Easy

If $^n{P_5} = 9 \times {^{n - 1}}{P_4}$,then the value of $n$ is

A
$6$
B
$8$
C
$5$
D
$9$

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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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If $^n{P_5} = 9 \times {^{n - 1}}{P_4}$,then the value of $n$ is

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Find the number of arrangements of the letters of the word $INDEPENDENCE$. In how many of these arrangements do all the vowels always occur together?

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