Question

$18$ guests have to be seated, half on each side of a long table. $4$ particular guests desire to seat on one particular side and $3$ on the other side. Then the number of ways in which the seating arrangement can be made is

• A. $(9!{)}^2$
• B. ${}^{11}{C}_4\cdot (9!{)}^2$
• C. ${}^{11}{C}_3\cdot (9!{)}^2$
• D. ${}^{11}{C}_5\cdot (9!{)}^2$

Answer 1

The correct option is D ${}^{11}{C}_5\cdot (9!{)}^2$

Total number of guests $=18$
Arrangement should be such that $4$ particular guests should be on one side and $3$ on other side.
So, $18-4-3=11$ seats left for other guests.
For particular side, say $S,$ four seats have been occupied. The remaining $5$ can be occupied in ${}^{11}{C}_5$ ways.
The remaining $6$ people out of $11$ automatically gets seats on the other side.
But the people in side $S$ can also be permuted in $9!$ ways.
Similarly, the people in the other side can also be permuted in $9!$ ways.
Hence, required number of arrangements $={}^{11}{C}_5\cdot {}^6{C}_6\cdot (9!{)}^2={}^{11}{C}_5\cdot (9!{)}^2$