If {}^n C_{r-1}=330,{}^n C_r=462,and {}^n C_{ | vedclass

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

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$5$

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(C) Given,${}^n C_{r-1}=330$,${}^n C_r=462$,and ${}^n C_{r+1}=462$. We know that $\frac{{}^n C_{r+1}}{{}^n C_r} = \frac{n-r}{r+1}$. Since ${}^n C_{r+1} = {}^n C_r = 462$,we have $\frac{n-r}{r+1} = 1$,which implies $n-r = r+1$,or $n = 2r+1$. Next,consider the ratio $\frac{{}^n C_r}{{}^n C_{r-1}} = \frac{462}{330} = \frac{7}{5}$. Using the formula $\frac{{}^n C_r}{{}^n C_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{n-r+1}{r} = \frac{7}{5}$. Substituting $n = 2r+1$ into this equation: $\frac{(2r+1)-r+1}{r} = \frac{7}{5} \Rightarrow \frac{r+2}{r} = \frac{7}{5}$. $5(r+2) = 7r$ $\Rightarrow 5r+10 = 7r$ $\Rightarrow 2r = 10$ $\Rightarrow r = 5$.

If ${}^n C_{r-1}=330$,${}^n C_r=462$,and ${}^n C_{r+1}=462$,then $r$ is equal to

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