(C) Given the ratio $^nP_4 : ^nP_5 = 1 : 2$. Using the formula $^nP_r = \frac{n!}{(n-r)!}$,we have: $\frac{n!}{(n-4)!} \div \frac{n!}{(n-5)!} = \frac{

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(C) Given the ratio $^nP_4 : ^nP_5 = 1 : 2$. Using the formula $^nP_r = \frac{n!}{(n-r)!}$,we have: $\frac{n!}{(n-4)!} \div \frac{n!}{(n-5)!} = \frac{1}{2}$ $\frac{n!}{(n-4)!} \times \frac{(n-5)!}{n!} = \frac{1}{2}$ $\frac{(n-5)!}{(n-4)(n-5)!} = \frac{1}{2}$ $\frac{1}{n-4} = \frac{1}{2}$ $n - 4 = 2$ $n = 6$

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

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If $^nP_4 : ^nP_5 = 1 : 2$,then $n = \dots$

A
$4$
B
$5$
C
$6$
D
$7$

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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 $^nP_4 : ^nP_5 = 1 : 2$,then $n = \dots$

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