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, the seating arrangement can be made?

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

Answer 1

The Correct Option is B

The problem involves arranging guests around a long table, with specific constraints on which side certain guests can sit. Let's solve the problem step-by-step: 

1. There are 20 guests in total. We need to arrange them such that 10 guests sit on each side of the table.
2. Three particular guests wish to sit on one specific side of the table. Let’s call this Side A. These guests must be included in the 10 guests on Side A.
3. Five other guests wish to sit on the opposite side, which we'll call Side B. These must be part of the 10 guests on Side B.
4. Since we need to fill seats on both sides, first consider Side A:
5. We've already placed 3 guests on Side A. Therefore, we need to choose \(10 - 3 = 7\) more guests to fill Side A.
6. There are \(20 - 3 - 5 = 12\) remaining guests who can be chosen to sit on either side. We choose any 7 of these 12 guests to sit on Side A, calculated as follows:

\({}\binom{12}{7} = \frac{12!}{7! \times 5!}\)

1. Next, Side B automatically gets the remaining 5 guests from those 12, plus the 5 guests who want to sit on Side B, for which we don’t need to calculate combinations since it’s by requirement. Thus, this arrangement of choosing guests is valid.
2. Now, calculate the ways to arrange the guests:
3. To seat the 10 guests on each side:

\(10!\\)ways for Side A.
\(10!\\)ways for Side B.

1. Combine these steps to get the total number of seating arrangements:

The final calculation for arranging all guests is:

\({}\binom{12}{7} \times 10! \times 10! = \frac{12!}{7! \times 5!} \times 10! \times 10!\)

The correct number of ways to seat the guests considering all constraints is:

\({}12! \times \frac{10!}{7!} \times \frac{10!}{5!}\)

Therefore, the correct answer is:

\(12! \times \frac{10!}{7!} \times \frac{10!}{5!}\)