Question

Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side. Determine n if the number of ways in which the seating arrangements can be made $\frac{n!}{5!6!}$ .$9!9!$

Answer 1

As information is given that $4$ particular guests has to sit on one side and other three also has to sit on another side.

Now, total $=18$

left $=18-7$

$=11$

So, total remaining $\left(11\right)$ left has $6$ seats on one side to sit and $5$ has $5$ sit on other side.

So, $={}^{11}{C}_6\times {}^5{C}_5$

where $9$ on each side can sit in $9!$ ways on each side