Browse Questions

# Determine $n$ if $2nC_3 : nC_3=11 : 1$

$\begin{array}{1 1}(A)\;4\\(B)\;6\\(C)\;8\\(D)\;10\end{array}$

Can you answer this question?

Toolbox:
• $nC_r=\bigg[\large\frac{n(n-1).......(n-r+1)}{1.2.3....n}\bigg]$
$\large\frac{2n(2n-1)(2n-2)}{1.2.3}/{\large\frac{n(n-1)(n-2)}{1.2.3}}=\frac{11}{1}$
$\large\frac{2n(2n-1)(2n-2)}{1.2.3}\times{\large\frac{6}{n(n-1)(n-2)}}=\frac{11}{1}$
$\large\frac{4n(n-1)(2n-1)}{1}\times \frac{1}{n(n-1)(n-2)}=\frac{11}{1}$
$\large\frac{4(2n-1)}{n-2}=\frac{11}{1}$
$4(2n-1)=11(n-2)$
$8n-4=11n-22$
$8n-11n=-22+4$
$-3n=-18$
$3n=18$
$n=6$
Hence (B) is the correct answer.
answered May 14, 2014