(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

Easy

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

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

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Class 11

Mathematics Questions

Chapter

Permutation and Combination

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

While dialing a telephone number,an old man forgets the last two digits,remembering only that these are different. If he dials the last two digits at random,what is the probability that the number is dialed correctly?

Easy

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${ }^{15} P_8 = A + 8 \cdot { }^{14} P_7 \Rightarrow A = $

DifficultTS EAMCET 2011

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If the number of words,with or without meaning,which can be made using all the letters of the word $MATHEMATICS$ in which $C$ and $S$ do not come together,is $(6 !) k$,then $k$ is equal to $............$.

DifficultJEE MAIN 2023

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

Medium

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The number of arrangements of the letters of the word $SALOON$,if the two $O$'s do not come together,is

Easy

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

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While dialing a telephone number,an old man forgets the last two digits,remembering only that these are different. If he dials the last two digits at random,what is the probability that the number is dialed correctly?

${ }^{15} P_8 = A + 8 \cdot { }^{14} P_7 \Rightarrow A = $

If the number of words,with or without meaning,which can be made using all the letters of the word $MATHEMATICS$ in which $C$ and $S$ do not come together,is $(6 !) k$,then $k$ is equal to $............$.

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