If 2 times {}^nC_5 = 9 times {}^{n-2}C_5,then | vedclass

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(B) Given the equation: $2 	imes {}^nC_5 = 9 	imes {}^{n-2}C_5$Using the formula ${}^nC_r = \frac{n!}{r!(n-r)!}$,we have:$2 	imes \frac{n!}{5!(n-5)

Vedclass · Accepted Answer

$10$

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(B) Given the equation: $2 	imes {}^nC_5 = 9 	imes {}^{n-2}C_5$Using the formula ${}^nC_r = \frac{n!}{r!(n-r)!}$,we have:$2 	imes \frac{n!}{5!(n-5)!} = 9 	imes \frac{(n-2)!}{5!(n-2-5)!}$$2 	imes \frac{n!}{5!(n-5)!} = 9 	imes \frac{(n-2)!}{5!(n-7)!}$Cancel $5!$ from both sides:$2 	imes \frac{n(n-1)(n-2)!}{(n-5)(n-6)(n-7)!} = 9 	imes \frac{(n-2)!}{(n-7)!}$Divide both sides by $(n-2)!$ and multiply by $(n-7)!$:$2 	imes \frac{n(n-1)}{(n-5)(n-6)} = 9$$2(n^2 - n) = 9(n^2 - 11n + 30)$$2n^2 - 2n = 9n^2 - 99n + 270$$7n^2 - 97n + 270 = 0$Solving the quadratic equation $7n^2 - 97n + 270 = 0$:$7n^2 - 70n - 27n + 270 = 0$$7n(n - 10) - 27(n - 10) = 0$$(7n - 27)(n - 10) = 0$Since $n$ must be an integer and $n \ge 7$ (for ${}^{n-2}C_5$ to be defined),we get $n = 10$.

If $2 \times {}^nC_5 = 9 \times {}^{n-2}C_5$,then the value of $n$ will be

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