Ans: (C)
Hint : \(\frac{\lfloor n}{|r-1| n-r+1}=36 \ldots \ldots(1)\)
\(\frac{\underline{\ln }}{\underline{\mathrm{r} \mid \mathrm{n}-\mathrm{r}}}=84 \ldots \ldots \ldots \ldots(2)\)
\(\frac{\ln }{|r+1| n-r-1}=126\)
\((1) \div(2)\) gives \(\frac{r}{n-r+1}=\frac{36}{84} \Rightarrow 84 r=36 n-36 r+36\) or \(120 r=36 n+36\)
(2) \(\div\) (3) gives \(\frac{r+1}{n-r}=\frac{84}{126} \Rightarrow 126 r+126=84 n-84 r\) or \(210 r=84 n-126\).
Solving \((4)\) and \((5) n=9, r=3\)
So \({ }^{n} \mathrm{C}_{8}={ }^{9} \mathrm{C}_{8}=9\)
