Determine $n$ if
$^{2 n} C_{3}:\,^{n} C_{3}=12: 1$
Determine $n$ if
$^{2 n} C_{3}:\,^{n} C_{3}=12: 1$
$\frac{{^{2n}{C_3}}}{{^n{C_3}}} = \frac{{12}}{1}$
$\Rightarrow \frac{(2 n) !}{3 !(2 n-3) !} \times \frac{3 !(n-3) !}{n !}=\frac{12}{1}$
$\Rightarrow \frac{(2 n)(2 n-1)(2 n-2)(2 n-3) !}{(2 n-3) !} \times \frac{(n-3) !}{n(n-1)(n-2)(n-3) !}=12$
$\Rightarrow \frac{2(2 n-1)(2 n-2)}{(n-1)(n-2)}=12$
$\Rightarrow \frac{4(2 n-1)(n-1)}{(n-1)(n-2)}=12$
$\Rightarrow \frac{(2 n-1)}{(n-2)}=3$
$\Rightarrow 2 n-1=3(n-2)$
$\Rightarrow 2 n-1=3 n-6$
$\Rightarrow 3 n-2 n=-1+6$
$\Rightarrow n=5$
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- [KVPY 2017]
$^n{C_r} + {2^n}{C_{r - 1}}{ + ^n}{C_{r - 2}} = $
$^n{C_r} + {2^n}{C_{r - 1}}{ + ^n}{C_{r - 2}} = $