if ⁿ⁺¹C_r+1 : ⁿC_r : ^n-1C_r-1=11:6:3 .FIND nr.

since 
$$\begin{gathered} \frac{{}^{n+1}C_{r+1}}{{}^{n}C_r}=\frac{(n+1)!}{(n-r)!*(r+1)!}*\frac{(n-r)!*r!}{n!} \\ \frac{{}^{n+1}C_{r+1}}{{}^{n}C_r}=\frac{(n+1)n!}{(r+1).r!}*\frac{r!}{n!} \\ \frac{{}^{n+1}C_{r+1}}{{}^{n}C_r}=\frac{n+1}{r+1} .........\left(1\right) \\ similarly: \\ \frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=\frac{n}{r} .........\left(2\right) \end{gathered}$$
therefore from the given equation
$$\begin{gathered} \frac{{}^{n+1}C_{r+1}}{{}^{n}C_r}=\frac{n+1}{r+1}=\frac{11}{6} \\ \frac{n+1}{r+1}=\frac{11}{6} \\ 6n+6=11r+11 \\ 6n-11r=5 ......\left(3\right) \end{gathered}$$
$$\begin{gathered} \frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=\frac{n}{r}=\frac{6}{3} \\ \frac{n}{r}=2 \\ \Rightarrow n=2r .....\left(4\right) \end{gathered}$$
from eq(3) and eq(4):
$$\begin{gathered} 6*2r-11r=5 \\ 12r-11r=5 \\ r=5 and n=2*5=10 \end{gathered}$$
thus n = 10 and r =5

hope this helps you