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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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Find $r$ if $5P_r=2^6P_{r-1}$

$\begin{array}{1 1}(A)\;2\\(B)\;3\\(C)\;4\\(D)\;5\end{array} $

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1 Answer

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  • $nP_r=\large\frac{n!}{(n-r)!}$
$5P_r=2^6P_{r-1}$
$\large\frac{5!}{(5-r)!}=$$2\times \large\frac{6!}{[6-(r-1)]!}$
$\Rightarrow 2\times \large\frac{6!}{(7-r)!}$
Or $\large\frac{5!}{(5-r)!}=\frac{2\times 6\times 5}{(7-r)(6-r)(5-r)!}$
$1=\large\frac{12}{(7-r)(6-r)}$
$(7-r)(6-r)=12$
$42-13r+r^2=12$
$r^2-13r+30=0$
$(r-10)(r-3)=0$
$r=3,10$
$r\neq 10$ Since r cannot be greater than n
$r=3$
Hence (B) is the correct answer.
answered May 13, 2014 by sreemathi.v
 

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