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Determine $n$ if $2nC_3 : nC_2=12 : 1$

$\begin{array}{1 1}(A)\;10\\(B)\;9\\(C)\;4\\(D)\;5\end{array} $

1 Answer

$nC_r=\bigg[\large\frac{n(n-1)....(n-r+1)}{1.2.3......n}\bigg]$
$\Rightarrow \large\frac{2n(2n-1)(2n-2)}{1.2.3}/\large\frac{n(n-1)}{1.2}=\frac{12}{1}$
(or) $\large\frac{2n(2n-1).2(n-1)}{6}\times \frac{2}{n(n-1)}=\frac{12}{1}$
$\large\frac{4n(2n-1)(n-1)}{3}\times \frac{1}{n(n-1)}=$$12$
$\large\frac{4(2n-1)}{3}$$=12$
$(2n-1)=\large\frac{12\times 3}{4}$
$2n-1=9$
$2n=10$
$n=5$
Hence (D) is the correct answer.
answered May 14, 2014 by sreemathi.v
 

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