Question

20 guests have to sit half on each side of a long table. Three particular guests desire to sit on a particular side, five others on the other side. In how many ways can the seating arrangement be made?

• A. \( 10! \times \dfrac{10!}{7!} \times \dfrac{10!}{5!} \)
• B. \( 12! \times \dfrac{10!}{7!} \times \dfrac{10!}{5!} \)
• C. \( 20! \times 13! \times 15! \)
• D. \( 20! \times \dfrac{7!}{3!} \)

Hint:

Split the 12 unrestricted guests between the two sides first, then arrange each side of 10.

Answer 1

The Correct Option is B

Step 1: Split the 20 seats by side.
The table seats 10 guests on each side, so there are two groups of 10 distinct seats.

Step 2: Place the guests with fixed sides.
The 3 guests who want a particular side must be among that side's 10 seats, and the 5 guests who want the other side must be among that side's 10 seats.
This leaves \( 20 - 3 - 5 = 12 \) unrestricted guests to fill the \( 10 - 3 = 7 \) remaining seats on the first side and the \( 10 - 5 = 5 \) remaining seats on the second side.

Step 3: Choose which guests fill which side.
Choose 7 of the 12 remaining guests for the first side in \( \binom{12}{7} = \dfrac{12!}{7!\,5!} \) ways; the other 5 automatically go to the second side.

Step 4: Arrange the guests within each side.
The 10 seats on each side are distinct positions, so the 10 guests on the first side can be arranged in \( 10! \) ways, and the 10 on the second side in another \( 10! \) ways.

Final Answer:
Total ways \( = \dfrac{12!}{7!\,5!} \times 10! \times 10! = 12! \times \dfrac{10!}{7!} \times \dfrac{10!}{5!} \). \[ \boxed{12! \times \dfrac{10!}{7!} \times \dfrac{10!}{5!}} \]