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# If $^nC_{r-1}=36$, $^nC_r=84$ and $^nC_{r+1}=126$, then $r$ = ?

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A)
$\large\frac{^nC_r}{^nC_{r-1}}=\frac{84}{36}$
$\Rightarrow\:\large\frac{n-r+1}{r}=\frac{7}{3}$
$\Rightarrow\:3n-10r+3=0$..........(i)
$\large\frac{^nC_{r+1}}{^nC_{r}}=\frac{126}{84}$
$\Rightarrow\:\large\frac{n-r}{r+1}=\frac{21}{14}$
$\Rightarrow\:14n-35r-21=0$..........(ii)
Solving (i) and (ii) we get $n=9 \:and\:r=3$