Q44 / 160 · TS EAMCET 2013 · Mathematics
If ${ }^n C_{r-1}=330,{ }^n C_r=462$, and ${ }^n C_{r+1}=462$, then $r$ is equal to
- A$3$
- B$4$
- Correct answerC$5$
- D$6$
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
$$
\begin{aligned}
& & 22 r & =110 \\
\therefore & & r & =5
\end{aligned}
$$
Hence, the correct answer is option: 'C'.