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

If ${ }^n C_{r-1}=330,{ }^n C_r=462$, and ${ }^n C_{r+1}=462$, then $r$ is equal to
  1. $3$
  2. $4$
  3. $5$
  4. $6$

Solution

Given, $ { }^n C_{r-1}=330,{ }^n C_r=462 $ and $\quad{ }^n C_{r+1}=462$ Now, $\quad \frac{{ }^n C_{r+1}}{{ }^n C_r}=1$ $ \begin{aligned} & \Rightarrow \quad \frac{\frac{n !}{(r+1) !(n-r-1) !}}{\frac{n !}{r !(n-r) !}}=1 \\ & \Rightarrow \quad \frac{r !(n-r)(n-r-1) !}{(r+1) r !(n-r-1) !}=1 \\ & \Rightarrow \quad \frac{n-r}{r+1}=1 \Rightarrow n-2 r=1 \end{aligned} $ Again, $\quad \frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{462}{330}=\frac{77}{55}$ $ \Rightarrow \quad \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r-1) !(n-r+1) !}}=\frac{77}{55} $ $ \begin{array}{ll} \Rightarrow & \frac{(r-1) !(n-r+1)(n-r) !}{r(r-1) !(n-r) !}=\frac{77}{55} \\ \Rightarrow & 55 n-132 r+55=0 \end{array} $ From Eqs. (i) and (ii), we get $ \begin{aligned} & & 22 r & =110 \\ \therefore & & r & =5 \end{aligned} $

Asked in: AP EAMCET 2013

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