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

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(B) Given the equation: $2 	imes \frac{n!}{5!(n-5)!} = 9 	imes \frac{(n-2)!}{5!(n-7)!}$ Cancel $5!$ from both sides: $2 	imes \frac{n!}{(n-5)!} = 9

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$10$

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(B) Given the equation: $2 	imes \frac{n!}{5!(n-5)!} = 9 	imes \frac{(n-2)!}{5!(n-7)!}$ Cancel $5!$ from both sides: $2 	imes \frac{n!}{(n-5)!} = 9 	imes \frac{(n-2)!}{(n-7)!}$ Expand the factorials: $2 	imes n(n-1)(n-2)! 	imes \frac{1}{(n-5)(n-6)(n-7)!} = 9 	imes \frac{(n-2)!}{(n-7)!}$ 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: $7n^2 - 70n - 27n + 270 = 0$ $7n(n - 10) - 27(n - 10) = 0$ $(7n - 27)(n - 10) = 0$ Since $n$ must be an integer,$n = 10$.

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

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