If ^{2n}C_3 : ^nC_2 = 44:3,then for which of | vedclass

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(B) Given the ratio: $\frac{^{2n}C_3}{^nC_2} = \frac{44}{3}$.Expanding the combinations: $\frac{\frac{(2n)(2n-1)(2n-2)}{3 	imes 2 	imes 1}}{\frac{n(

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$r = 4$

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(B) Given the ratio: $\frac{^{2n}C_3}{^nC_2} = \frac{44}{3}$.Expanding the combinations: $\frac{\frac{(2n)(2n-1)(2n-2)}{3 	imes 2 	imes 1}}{\frac{n(n-1)}{2 	imes 1}} = \frac{44}{3}$.Simplifying the expression: $\frac{(2n)(2n-1) \cdot 2(n-1)}{6} \cdot \frac{2}{n(n-1)} = \frac{44}{3}$.$\frac{4n(2n-1)(n-1)}{6n(n-1)} = \frac{44}{3} \Rightarrow \frac{2(2n-1)}{3} = \frac{44}{3}$.$2(2n-1) = 44 \Rightarrow 2n-1 = 22 \Rightarrow 2n = 23$ (Wait,re-evaluating: $\frac{2n(2n-1)(2n-2)}{6} \cdot \frac{2}{n(n-1)} = \frac{44}{3} \Rightarrow \frac{4n(2n-1)(n-1)}{3n(n-1)} = \frac{44}{3} \Rightarrow 4(2n-1) = 44 \Rightarrow 2n-1 = 11 \Rightarrow 2n = 12 \Rightarrow n = 6$).Now,we need $^6C_r = 15$. Since $^6C_2 = \frac{6 	imes 5}{2} = 15$ and $^6C_4 = ^6C_2 = 15$,the possible values for $r$ are $2$ or $4$. Comparing with the options,$r = 4$ is correct.

If $^{2n}C_3 : ^nC_2 = 44:3$,then for which of the following values of $r$,the value of $^nC_r$ will be $15$?

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