Determine n if ^{2n}C_{3} : ^{n}C_{3} = 11 : | vedclass

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(B) Given $\frac{^{2n}C_{3}}{^{n}C_{3}} = \frac{11}{1}$.Using the formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:$\frac{(2n)!}{3!(2n-3)!} 	imes \f

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Given $\frac{^{2n}C_{3}}{^{n}C_{3}} = \frac{11}{1}$.Using the formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:$\frac{(2n)!}{3!(2n-3)!} 	imes \frac{3!(n-3)!}{n!} = 11$$\Rightarrow \frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} 	imes \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 11$$\Rightarrow \frac{(2n)(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 11$$\Rightarrow \frac{2(2n-1) \cdot 2}{n-2} = 11$$\Rightarrow \frac{4(2n-1)}{n-2} = 11$$\Rightarrow 8n - 4 = 11n - 22$$\Rightarrow 3n = 18$$\Rightarrow n = 6$.

Determine $n$ if $^{2n}C_{3} : ^{n}C_{3} = 11 : 1$.

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