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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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Find $r$ if $5P_r=6P_{r-1}$

$\begin{array}{1 1}(A)\;4\\(B)\;5\\(C)\;6\\(D)\;8\end{array} $

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1 Answer

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  • $nP_r=\large\frac{n!}{(n-r)!}$
$\large\frac{5!}{(5-r)!}=\frac{6!}{[6-(r-1)]!}$
$\Rightarrow \large\frac{6!}{(7-r)!}$
$\large\frac{5!}{(5-r)!}=\frac{6.5!}{(7-r)(6-r)(5-r)!}$
$\therefore 1=\large\frac{6}{(7-r)(6-r)}$
(Or) $(7-r)(6-r)=6$
(Or) $42-13r+r^2=6$
(Or) $r^2-3r+36=0$
(Or) $(r-9)(r-4)=0$
$r=9,4$.$r=4$
Since $r\neq 9$,$r$ cannot greater than 4
Hence (A) is the correct answer.
answered May 13, 2014 by sreemathi.v
 

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