If 12 cdot {}^{n}C_{2} = {}^{2n}C_{3},find n. | vedclass

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(B) Given equation: $12 \cdot {}^{n}C_{2} = {}^{2n}C_{3}$.Using the formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:$12 \cdot \frac{n!}{2!(n-2)!}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Given equation: $12 \cdot {}^{n}C_{2} = {}^{2n}C_{3}$.Using the formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:$12 \cdot \frac{n!}{2!(n-2)!} = \frac{(2n)!}{3!(2n-3)!}$$12 \cdot \frac{n(n-1)}{2} = \frac{(2n)(2n-1)(2n-2)}{6}$$6n(n-1) = \frac{2n(2n-1) \cdot 2(n-1)}{6}$$6n(n-1) = \frac{4n(2n-1)(n-1)}{6}$Dividing both sides by $2n(n-1)$ (assuming $n > 1$):$3 = \frac{2(2n-1)}{6}$$18 = 4n - 2$$20 = 4n$$n = 5$.

If $12 \cdot {}^{n}C_{2} = {}^{2n}C_{3}$,find $n$.

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