Question

5 boys and 5 girls have to sit around a table. The number of ways in which all of them can sit so that no two boys and no two girls are together is

• A. 14400
• B. 2880
• C. 576
• D. 625

Hint:

For alternating arrangements in a circle (e.g., n men and n women), first arrange one group in \((n-1)!\) ways. Then, arrange the second group in the \(n\) spaces created, which can be done in \(n!\) ways. The total ways are \((n-1)! \times n!\).

Answer 1

The Correct Option is B

The condition that no two boys and no two girls sit together means that the seating arrangement must be alternating between boys and girls (e.g., B-G-B-G-...).
This is a circular permutation problem. We first fix the positions for one group (either boys or girls) and then arrange the other group in the gaps.
Step 1: Arrange the 5 boys around the circular table.
The number of ways to arrange \(n\) distinct objects in a circle is \((n-1)!\).
So, the number of ways to seat the 5 boys is \((5-1)! = 4!\).
\(4! = 4 \times 3 \times 2 \times 1 = 24\).
Step 2: Arrange the 5 girls in the seats between the boys.
Once the boys are seated, they create 5 distinct empty seats between them.
Since the positions are now fixed relative to the boys, this part of the problem is a linear permutation. We need to arrange 5 girls in 5 specific chairs.
The number of ways to arrange 5 girls in these 5 seats is \(5!\).
\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Step 3: Calculate the total number of ways.
The total number of arrangements is the product of the number of ways from Step 1 and Step 2.
Total ways = (Ways to arrange boys) \(\times\) (Ways to arrange girls) = \(4! \times 5!\).
Total ways = \(24 \times 120 = 2880\).