Question

If ${}^{n+1}{C}_{r+1}{:}^n{C}_r{:}^{n-1}{C}_{r-1}=11:6:3,nr$ is equal to

Answer 1

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

Therefore,

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

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

$\therefore \frac{n+1}{r+1}=\frac{11}{6}.....\left(1\right)$

Similarly,

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

$n=2r.....\left(2\right)$

Now from ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\frac{2r+1}{r+1}=\frac{11}{6}$

$\Rightarrow 6(2r+1)=11(r+1)$

$\Rightarrow 12r+6=11r+11$

$\Rightarrow r=5$

Substituting the value of $r$ in ${eq}^n\left(2\right)$, we have

$n=2r=10$

$\therefore nr=5\times 10=50$

Hence the value of $nr$ is $50$.