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If $^nP_r=720^nC_r$, then r=?

$\begin{array}{1 1} 4 \\ 5 \\ 6 \\ 7 \end{array}$

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$^nP_r=\large\frac{n!}{(n-r)!}$$\:\:and\:\:^nC_r=\large\frac{n!}{(n-r)!.r!}$
Given: $ ^nP_r=720.^nC_r$
$\Rightarrow\:\large\frac{n!}{(n-r)!}$=$720.\large\frac{n!}{(n-r)!.r!}$
$\Rightarrow\:r!=720\:\Rightarrow\:r=6$
answered Sep 6, 2013 by rvidyagovindarajan_1
 

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