Determine n if ^{2n}C_{3} : ^{n}C_{3} = 12 : | vedclass

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(A) Given $\frac{^{2n}C_{3}}{^{n}C_{3}} = \frac{12}{1}$.Using the formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:$\frac{(2n)!}{3!(2n-3)!} 	imes \f

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given $\frac{^{2n}C_{3}}{^{n}C_{3}} = \frac{12}{1}$.Using the formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:$\frac{(2n)!}{3!(2n-3)!} 	imes \frac{3!(n-3)!}{n!} = 12$.Simplifying the factorials:$\frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} 	imes \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 12$.$\frac{(2n)(2n-1)(2(n-1))}{n(n-1)(n-2)} = 12$.$\frac{2(2n-1)(2)}{n-2} = 12$.$\frac{4(2n-1)}{n-2} = 12$.$\frac{2n-1}{n-2} = 3$.$2n - 1 = 3(n - 2)$.$2n - 1 = 3n - 6$.$n = 5$.

Determine $n$ if $^{2n}C_{3} : ^{n}C_{3} = 12 : 1$.

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