If {}^{2n}C_3 : {}^{n}C_3 = 10 : 1,then the r | vedclass

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(D) Given $\frac{{}^{2n}C_3}{{}^{n}C_3} = 10$. Using the formula ${}^{n}C_r = \frac{n!}{r!(n-r)!}$,we have: $\frac{\frac{(2n)(2n-1)(2n-2)}{3 	imes 2

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$2 : 1$

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(D) Given $\frac{{}^{2n}C_3}{{}^{n}C_3} = 10$. Using the formula ${}^{n}C_r = \frac{n!}{r!(n-r)!}$,we have: $\frac{\frac{(2n)(2n-1)(2n-2)}{3 	imes 2 	imes 1}}{\frac{n(n-1)(n-2)}{3 	imes 2 	imes 1}} = 10$ $\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 10$ $\frac{2(2n-1) \cdot 2(n-1)}{(n-1)(n-2)} = 10$ $\frac{4(2n-1)}{n-2} = 10$ $8n - 4 = 10n - 20$ $2n = 16 \Rightarrow n = 8$. Now,substitute $n = 8$ into the ratio $(n^2 + 3n) : (n^2 - 3n + 4)$: $n^2 + 3n = 8^2 + 3(8) = 64 + 24 = 88$ $n^2 - 3n + 4 = 8^2 - 3(8) + 4 = 64 - 24 + 4 = 44$ Ratio $= 88 : 44 = 2 : 1$.

If ${}^{2n}C_3 : {}^{n}C_3 = 10 : 1$,then the ratio $(n^2 + 3n) : (n^2 - 3n + 4)$ is

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