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# If $nC_{r-1}=36,nC_r=84$ and $nC_{r+1}=126$ then find $rC_2$

$\begin{array}{1 1}(A)\;1\\(B)\;2\\(C)\;4\\(D)\;3\end{array}$

Toolbox:
• $\large\frac{nC_r}{nC_{r-1}}=\frac{n!}{r!(n-r)!}\times \frac{(-1)!(n-r+1)!}{n!}$
• $\large\frac{nC_{r+1}}{nC_r}=\frac{n-(r+1)+1}{r+1}$
We have
$\large\frac{nC_r}{nC_{r-1}}=\frac{n!}{r!(n-r)!}\times \frac{(r-1)!(n-r+1)!}{n!}$
$\Rightarrow \large\frac{n-r+1}{r}$
Similarly
$\large\frac{nC_{r+1}}{nC_r}=\frac{n-(r+1)+1}{r+1}$
$\Rightarrow \large\frac{n-r}{r+1}$
$\therefore \large\frac{n-r+1}{r}=\frac{84}{36}$
$\Rightarrow \large\frac{7}{3}$
$\large\frac{n-r}{r+1}=\frac{126}{84}$
$\Rightarrow \large\frac{3}{2}$
$\large\frac{n-r+1}{r}=\frac{7}{3}$
$3n-3r+1=7r$
$3n-3r-7r+1=0$
$3n-10r+3=0$------(1)
$\large\frac{n-r}{r+1}=\frac{3}{2}$
$2(n-r)=3(r+1)$
$2n-2r=3r+3$
$2n-2r-3r-3=0$
$2n-5r-3=0$------(2)
$3n-10r+3$
$4n\pm 10r\pm 6$
____________________
-n+9
$-n+9=0$
$-n=-9$
$n=9$
$3(9)-10(r)+3=0$
$27-10r+3=0$
$-10r=-30$
$r=3$
$rC_2=3C_2$
$\Rightarrow \large\frac{3!}{2!\times 1!}$
$\Rightarrow \large\frac{3\times 2\times 1}{2\times 1}$
$\Rightarrow 3$
Hence (D) is the correct answer.