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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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Find $n$ if $^{n-1}P_3 : ^nP_4=1 : 9$

$\begin{array}{1 1}(A)\;6\\(B)\;7\\(C)\;8\\(D)\;9\end{array} $

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1 Answer

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  • $^nP_r=\large\frac{n!}{(n-r)!}$
$\large\frac{{n-1}P_3}{^nP_4}=\frac{1}{9}$
$\large\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{9}$
$\Rightarrow \large\frac{1}{n}=\frac{1}{9}$
$\Rightarrow n=9$
Hence (D) is the correct answer.
answered May 13, 2014 by sreemathi.v
 

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