If ${ }^n C_{r-1}=36,{ }^n C_r=84$ and ${ }^n C_{r+1}=126$, then
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If ${ }^n C_{r-1}=36,{ }^n C_r=84$ and ${ }^n C_{r+1}=126$, then

(A) $n=8, r=4$

(B) $n=9, r=3$

(C) $n=7, r=5$

(D) None of these

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SOLUTION —

$\begin{array}{l}\text { Now, } \frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{84}{36} \text { and } \frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{126}{84} \\\Rightarrow \quad \frac{n-r+1}{r}=\frac{7}{3} \text { and } \frac{n-r}{r+1}=\frac{3}{2} \\\Rightarrow \quad 3 n-10 r+3=0 \text { and } 2 n-5 r-3=0 \\\Rightarrow \quad r=3 \text { and } n=9 \\\end{array}$

So, The correct option will be (B). 

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