^{2n}C_4 : ^nC_3 = 99 : 4 Rightarrow n = | vedclass

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(B) Given the ratio: $\frac{^{2n}C_4}{^nC_3} = \frac{99}{4}$ Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$: $\frac{\frac{(2n)!}{4!(2n-4)!}}{\frac{n!

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Given the ratio: $\frac{^{2n}C_4}{^nC_3} = \frac{99}{4}$ Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$: $\frac{\frac{(2n)!}{4!(2n-4)!}}{\frac{n!}{3!(n-3)!}} = \frac{99}{4}$ $\frac{(2n)(2n-1)(2n-2)(2n-3)}{4 	imes 3 	imes 2 	imes 1} 	imes \frac{3 	imes 2 	imes 1}{n(n-1)(n-2)} = \frac{99}{4}$ $\frac{(2n)(2n-1) 	imes 2(n-1)(2n-3)}{4 	imes n(n-1)(n-2)} = \frac{99}{4}$ $\frac{2(2n-1)(2n-3)}{n-2} = 99$ $4(4n^2 - 6n - 2n + 3) = 99(n-2)$ $16n^2 - 32n + 12 = 99n - 198$ $16n^2 - 131n + 210 = 0$ $(n-6)(16n-35) = 0$ Since $n$ must be an integer,$n = 6$.

$^{2n}C_4 : ^nC_3 = 99 : 4 \Rightarrow n = $

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