If ²ⁿC₃ : ⁿC₂ = 44 : 3, find n.

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Solution

Given: $2 n_{C_{3}} : n_{C_{2}} = 44 : 3$
$\frac{2 n_{C_{3}}}{n_{C_{2}}} = \frac{44}{3} \Rightarrow \frac{2 n!}{3! (2 n - 3)!} \times \frac{2! (n - 2)!}{n!} = \frac{44}{3} \Rightarrow \frac{2 n (2 n - 1) (2 n - 2)}{3 n (n - 1)} = \frac{44}{3} \Rightarrow (2 n - 1) (2 n - 2) = 22 (n - 1) \Rightarrow 4 n^{2} - 6 n + 2 = 22 n - 22 \Rightarrow 4 n^{2} - 28 n + 24 = 0 \Rightarrow n^{2} - 7 n + 6 = 0 \Rightarrow n^{2} - 6 n - n + 6 = 0 \Rightarrow n (n - 6) - 1 (n - 6) = 0 \Rightarrow (n - 1) (n - 6) = 0$
$\Rightarrow n = 1$ or, $n = 6$

Now, $n = 1 \Rightarrow 2_{C_{3}} : 2_{C_{2}} = 44 : 3$
But, this is not possible.
∴ $n = 6$

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