Question

$\left(\begin{array}{c}n+1\\ r+1\end{array}\right):\left(\begin{array}{c}n\\ r\end{array}\right):\left(\begin{array}{c}n-1\\ r-1\end{array}\right)=11:6:3,,$ find $n$ & $r$.

Answer 1

From question we get,

${}^{n+1}{C}_{r+1}{:}^n{C}_r{:}^{n-1}{C}_{r-1}=11:6:3$

$\frac{{}^{n+1}{C}_{r+1}}{{}^n{C}_r}=\frac{(n+1)!}{(n-r)!(r+1)!}\times \frac{(n-r)!r!}{n!}$

$\frac{{}^{n+1}{C}_{r+1}}{{}^n{C}_r}=\frac{(n+1)n!}{(r+1).r!}\times \frac{r!}{n!}$

$\frac{{}^{n+1}{C}_{r+1}}{{}^n{C}_r}=\frac{n+1}{r+1}$ ----- ( 1 )

Similarly,

$\frac{{}^n{C}_r}{{}^{n-1}{C}_{r-1}}=\frac{n}{r}$ ------ ( 2 )

From the question,

$\frac{{}^{n+1}{C}_{r+1}}{{}^n{C}_r}=\frac{n+1}{r+1}=\frac{11}{6}$

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

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

$\Rightarrow$ $6n-11r=5$ ---- ( 3 )

$\Rightarrow$ $\frac{{}^n{C}_r}{{}^{n-1}{C}_{r-1}}=\frac{n}{r}=\frac{6}{3}$

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

$\Rightarrow$ $n=2r$ ------ ( 4 )

From equation ( 3 ) and equation ( 4 )

$\Rightarrow$ $6\times 2r-11r=5$

$\Rightarrow$ $12r-11r=5$

$\Rightarrow$ $r=5$ and $n=2\times 5=10$