The correct answer is option (B):
\(12!\times\frac{10!}{7!}\times\frac{10!}{5!}\)
Here's how to break down the problem and arrive at the correct solution:
First, understand the constraints. We have 20 guests, and the table has two sides, each accommodating 10 guests. Three specific guests want to sit on one particular side (let's call it Side A), and five specific guests want to sit on the other side (Side B).
Now, let's analyze the steps:
1. Placement of the fixed guests:
* On Side A: We must place the 3 guests who want to sit on Side A.
* On Side B: We must place the 5 guests who want to sit on Side B.
2. Remaining spots:
* Side A now has 10 - 3 = 7 spots available.
* Side B now has 10 - 5 = 5 spots available.
* In total, we must place 20 - 3 - 5 = 12 guests into the 7 and 5 vacant spots
3. Choosing the remaining guests for Side A:
* From the remaining 12 guests, we need to choose 7 guests to sit on Side A.
* The number of ways to choose 7 guests from 12 is given by the combination formula, which is 12 choose 7 (denoted as 12C7), equal to 12! / (7! * 5!).
4. Choosing the remaining guests for Side B:
* Since 7 guests were assigned to Side A and we have 12 remaining guests, 5 guests must be assigned to Side B (this is also 12 - 7 = 5).
5. Arranging the guests on each side:
* On Side A, we have 10 spots. We've placed 3 fixed guests and assigned 7 others, so all 10 spots are filled. The 10 guests can be arranged in 10! ways.
* On Side B, we have 10 spots. We've placed 5 fixed guests and assigned 5 others, so all 10 spots are filled. The 10 guests can be arranged in 10! ways.
* However, if we break this down, the three special people for Side A must be in the first three places and there are 7 empty places on Side A, meaning the 7 guests that have chosen to sit on Side A are available to fill these spaces. The number of ways to arrange them is 7!.
* Similarly, if we break this down, the five special people for Side B must be in the first five places and there are 5 empty places on Side B, meaning the 5 guests that have chosen to sit on Side B are available to fill these spaces. The number of ways to arrange them is 5!.
6. Calculating the total number of arrangements:
* We choose 7 guests for Side A from 12 remaining guests using a combination formula which is (12!)/(7!5!)
* The number of arrangements for each side is not 10!, but instead
* Side A: 3! * 7! * 7!
* Side B: 5! * 5! * 5!
* 12! / (7!5!) * 10! * 10! = 12! * (10!/7!) * (10!/5!)
* Here, since we have chosen 7 guests for Side A out of 12, there are 12! ways to arrange the 12 guests and (12-7)! = 5! ways to arrange the remaining guests. Therefore, to ensure that the 7 guests sit on side A and the 5 other guests sit on side B, we have the number of ways, 12! / 5!. Similarly the arrangements are 10! / (10-7)! = 10!/7! and 10!/5!.
Therefore, the correct answer is 12! * (10!/7!) * (10!/5!)