Question

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

• A. 20
• B. 30
• C. 40
• D. 5

Answer 1

Answer: (D)

5

$\frac{{}^{n+1}{C}_{r+1}}{{}^n{C}_r}=\frac{11}{6}$
$\frac{(n+1)!.(n-r)!.r!}{(n-r)!(r+1)!n!}=\frac{11}{6}$
$\frac{n+1}{r+1}=\frac{11}{6}$
$6n+6=11r+11$
$6n-11r=5$ ...(i)
and
$\frac{{}^n{C}_r}{{}^{n-1}{C}_{r-1}}=2$
$\frac{n!(n-r)!(r-1)!}{(n-1)!(n-r)!r!}=2$
$\frac{n}{r}=2$
$n=2r$ ...(ii)
Substituting in (i), we get
$6\left(2r\right)-11r=5$
$r=5$ and $n=10$