Question

Eighteen guests have to be seated, half on each side a long table. Four particular guest desires to sit on one particular side and three others on the other side. The number of ways in which the seating arrangement can be made is

• A. $9!\times 9!$
• B. ${}^{11}{C}_5\times 9!\times 9!$
• C. $\frac{11!}{5!}\times 9!\times 9!$
• D. ${}^{11}{C}_5$

Answer 1

The correct option is B ${}^{11}{C}_5\times 9!\times 9!$

Out of guests half i.e. to be seated on the side $A$ and rest $9$ on the side $B.$

Now out of $18$ guests, $4$ particular guests desire to sit – on one particular side say side $A$ and other $3$ on another side $B.$

Out of rest $18–4–3=11$ guests we can select $5$ more for side $A$and rest $6$ can be seated on side $B.$

Selection of $5$ out of $11$ can be done in ${{11}_C}_5$ways and $9$ guests on each side of the table can be seated in $9!\times 9!$ Ways.

Thus there are total${{11}_C}_5$ $\times 9!\times 9!$ Arrangements.