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 the other side. Determine the number of ways in which the sitting arrangement can be made.

• A. ${}^{11}{C}_5.(9!{)}^2$
• B. ${}^{12}{C}_4.(9!{)}^2$
• C. ${}^{11}{C}_5.(8!{)}^2$
• D. ${}^{12}{C}_4.(8!{)}^2$

Answer 1

Answer: (A)

${}^{11}{C}_5.(9!{)}^2$

$4+3$ guests have already selected the sides where they will be seated.

Out of the remaining $11$ guests, we have to select $5$ for one side, and the remaining will sit on the other side.
We can choose $5$ guest for a particular side in ${}^{11}{\text{C}}_5$ ways
Now $9$ guests in the particular side can be arranged in $9!$ ways and $9$ on the other side in $9!$ ways as well.
Hence the total number of ways $={}^{11}{\text{C}}_5.(9!{)}^2$