If 36 people are to be seated around a square table 9 each side then the probability that a particular child always sit between his/her parents on the same side of table is  $\frac{p}{q}$ where p and q are co – prime to each other then

Number of circular permutations of 36 people=35!

Number of positions of the particular child sitting on a particular side = 9

E = child sits in between the parents

chose the three (Parents, child )as one unit. 

no of persons =36-3+1=34 can be arranged in  circular manner in 33! ways.

Group of three can be arranged on a side in 7 ways , parents on either side of their child can be arranged in 2 ways.

$$\begin{gathered} n\left(S\right)=9\times 35! \\ \begin{array}{l}n\left(E\right)=7\times 33!\times 2\\ p\left(E\right)=\frac{7\times 33!\times 2}{9\times 35\times 34\times 33!}=\frac{1}{9\times 5\times 17}\end{array} \end{gathered}$$