If binom{2n}{3} : binom{n}{2} = 44 : 3 and bi | vedclass

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(B) Given $\binom{2n}{3} : \binom{n}{2} = 44 : 3$.Using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$,we have:$\frac{(2n)!}{3!(2n-3)!} 	imes \frac

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$4$

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(B) Given $\binom{2n}{3} : \binom{n}{2} = 44 : 3$.Using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$,we have:$\frac{(2n)!}{3!(2n-3)!} 	imes \frac{2!(n-2)!}{n!} = \frac{44}{3}$Expanding the factorials:$\frac{2n(2n-1)(2n-2)(2n-3)!}{6(2n-3)!} 	imes \frac{2(n-2)!}{n(n-1)(n-2)!} = \frac{44}{3}$$\frac{2n(2n-1) \cdot 2(n-1)}{6} 	imes \frac{2}{n(n-1)} = \frac{44}{3}$$\frac{4n(2n-1)(n-1)}{6n(n-1)} = \frac{44}{3}$$\frac{2(2n-1)}{3} = \frac{44}{3}$$2(2n-1) = 44 \implies 2n-1 = 22 \implies 2n = 23$ (Wait,re-evaluating the simplification).Correction: $\frac{2n(2n-1)(2n-2)}{6} 	imes \frac{2}{n(n-1)} = \frac{44}{3}$$\frac{2n(2n-1) \cdot 2(n-1)}{6} 	imes \frac{2}{n(n-1)} = \frac{4(2n-1)}{6} = \frac{2(2n-1)}{3} = \frac{44}{3}$$2(2n-1) = 44 \implies 2n-1 = 22 \implies 2n = 23$ (There is a slight error in the provided solution's arithmetic).Let's re-calculate: $\frac{2n(2n-1)(2n-2)}{6} \cdot \frac{2}{n(n-1)} = \frac{2n(2n-1) \cdot 2(n-1)}{6} \cdot \frac{2}{n(n-1)} = \frac{4(2n-1)}{6} = \frac{2(2n-1)}{3}$.If $\frac{2(2n-1)}{3} = \frac{44}{3}$,then $2n-1 = 22$,$n=11.5$. Checking the ratio again: $\binom{2n}{3} / \binom{n}{2} = \frac{2n(2n-1)(2n-2)}{6} / \frac{n(n-1)}{2} = \frac{2n(2n-1)2(n-1)}{6} \cdot \frac{2}{n(n-1)} = \frac{8n(2n-1)(n-1)}{6n(n-1)} = \frac{4(2n-1)}{3} = \frac{44}{3}$.$4(2n-1) = 44 \implies 2n-1 = 11 \implies 2n = 12 \implies n = 6$.Now,$\binom{6}{r} = 15$. Since $\binom{6}{2} = 15$ and $\binom{6}{4} = 15$,$r$ can be $2$ or $4$. Given the options,$r=4$ is correct.

If $\binom{2n}{3} : \binom{n}{2} = 44 : 3$ and $\binom{n}{r} = 15$,then $r = \dots$

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