If {}^nC_{r-1}=330,{}^nC_r=462,and {}^nC_{r+1 | vedclass

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(C) Given,${}^nC_{r-1}=330$,${}^nC_r=462$,and ${}^nC_{r+1}=462$. We know that $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1}$. Since ${}^nC_{r+1} = {}

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$5$

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(C) Given,${}^nC_{r-1}=330$,${}^nC_r=462$,and ${}^nC_{r+1}=462$. We know that $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1}$. Since ${}^nC_{r+1} = {}^nC_r = 462$,we have $\frac{462}{462} = 1$. Therefore,$\frac{n-r}{r+1} = 1 \implies n-r = r+1 \implies n = 2r+1$. Now,consider the ratio $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{462}{330}$. Using the formula $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{n-r+1}{r} = \frac{462}{330} = \frac{7}{5}$. Substituting $n = 2r+1$ into the equation: $\frac{(2r+1)-r+1}{r} = \frac{7}{5} \implies \frac{r+2}{r} = \frac{7}{5}$. $5(r+2) = 7r \implies 5r + 10 = 7r \implies 2r = 10 \implies r = 5$.

If ${}^nC_{r-1}=330$,${}^nC_r=462$,and ${}^nC_{r+1}=462$,then $r$ is equal to

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