Solve this

Solution:

Given: ${ }^{n+1} C_{r+1}:{ }^{n} C_{r}=11: 6$ and ${ }^{n} C_{r}:{ }^{n-1} C_{r-1}=6: 3$

To Find : $n \& r$

We use this property in this question:

$\left(\begin{array}{l}n \\ r\end{array}\right)=\frac{n}{r} \times\left(\begin{array}{l}n-1 \\ r-1\end{array}\right)$

${ }^{n+1} C_{r+1}:{ }^{n} C_{r}=11: 6$

$\Rightarrow \frac{\left(\begin{array}{l}n+1 \\ r+1\end{array}\right)}{\left(\begin{array}{l}n \\ r\end{array}\right)}=\frac{11}{6}$

$\Rightarrow \frac{\frac{(n+1)}{(r+1)} \times\left(\begin{array}{c}n \\ r\end{array}\right)}{\left(\begin{array}{l}n \\ r\end{array}\right)}=\frac{11}{6}$

$\Rightarrow \frac{(n+1)}{(r+1)}=\frac{11}{6}$

$\Rightarrow 6(n+1)=11(r+1)$

$\Rightarrow 6 n+6=11 r+11$

$\Rightarrow 6 n-11 r=5 \ldots(1)$

${ }^{n} C_{r}:{ }^{n-1} C_{r-1}=6: 3$

$\Rightarrow \frac{\left(\begin{array}{c}n \\ r\end{array}\right)}{\left(\begin{array}{c}n-1 \\ r-1\end{array}\right)}=\frac{6}{3}=2$

$\Rightarrow \frac{\frac{n}{r} \times\left(\begin{array}{l}n-1 \\ r-1\end{array}\right)}{\left(\begin{array}{c}n-1 \\ r-1\end{array}\right)}=2$

$\Rightarrow \frac{n}{r}=2$

$\Rightarrow \mathrm{n}=2 \mathrm{r} \ldots(2)$

Using equations 1 & 2 we get

$\Rightarrow 6(2 r)-11 r=5$

$\Rightarrow 12 r-11 r=5$

$\Rightarrow r=5$

$\Rightarrow n=2 \times 5$

$\Rightarrow n=10$

Ans: $n=10 \& r=5$