If ${^{2 n} C}_{3} : {^{n} C}_{2} = 44 : 3$ , find $n$ .

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Solution

${^{2 n} C}_{3} : {^{n} C}_{2} = 44 : 3$

$⟹ \frac{(2 n)!}{(2 n - 3)! \times 3!} \times \frac{(n - 2)! \times 2!}{n!} = \frac{44}{3}$
$⟹ \frac{2 n (2 n - 1) (2 n - 2)}{n (n - 1) \times 3} = \frac{44}{3}$
$⟹ \frac{2 (2 n - 1) (n - 1)}{n - 1} = 22$

$⟹ 2 n^{2} - 3 n + 1 = 11 n - 11$
$⟹ 2 n^{2} - 14 n + 12 = 0$
$⟹ (2 n - 2) (n - 6) = 0$
$⟹ n = 1, 6$
Since $^{1} C_{2}$ is not possible
$⟹∴ n = 6$

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^{2 n} C_{3}

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