Question
n का मान निकालिए, यदि $^{2n}C_3:\ ^nC_3 = 11 : 1$

Answer

$^{2n}C_3:\ ^nC_3 = 11 : 1$
$\Rightarrow \frac{2n!}{3! 2n! - 3!}: \frac{n!}{3!n!-3!} = 11 : 1$
$\Rightarrow \frac{2n(2 n-1)(2 n-2)}{3!}: \frac{n(n-1)(n-2)}{3!} = 11 : 1$
$\Rightarrow \frac{2 n(2 n-1)(2 n-2)}{n(n-1)(n-2)}=\frac{11}{1} $
$\Rightarrow \frac{4(2 n-1)}{n-2} = 11$
$\Rightarrow 8n - 4 = 11n - 22$
$\Rightarrow 3n = 18, \therefore n = 6$

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