Question

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

• A. $20$
• B. $30$
• C. $40$
• D. $50$

Answer 1

Answer: (D)

Given,
$\frac{{}^{n+1}{C}_{r+1}}{{}^n{C}_r}=\frac{11}{6}$ or $\frac{\frac{n+1}{r+1}{\times }^n{C}_r}{{}^n{C}_r}=\frac{11}{6}$
or $6n+6=11r+11$ or $6n-11r=\mathrm{5....}\left(1\right)$
Also,
$\frac{{}^n{C}_r}{{}^{n-1}{C}_{r-1}}=\frac{6}{3}$ or $\frac{\frac{n}{r}{\times }^{n-1}{C}_{r-1}}{{}^{n-1}{C}_{r-1}}=\frac{6}{3}$ or $n=2r....\left(2\right)$
From (1) and (2), $r=5$ and $n=10$,
$\therefore nr=50$