Question

7 boys and 5 girls are to be seated around a circular table such that no two girls sit together is?

• A. \(126(5!)^{2}\)
• B. \(720(5!)\)
• C. \(720(6!)\)
• D. 720

Answer 1

The Correct Option is A

The correct answer is (A) : 
First, we need to find the total number of ways to seat all 12 people around the circular table, which is (12-1)! = 11! since we can fix one person's position as a reference.
Next, we need to subtract the number of ways that two or more girls sit together. We can approach this by treating the five girls as a block and permuting them first, which can be done in 5! ways. 
Then we can insert this block of girls in the 8 spaces between the 7 boys or at the beginning or end of the line of boys, which gives us 9 positions to place the block of girls. Once the block of girls is placed, we can permute the 7 boys in 7! ways. Therefore, the total number of ways that two or more girls sit together is 5! × 9 × 7!
\(\therefore\) the number of ways that no two girls sit together is 11! - 5! × 9 × 7! = 126(5!)².

 

Answer 2

The correct answer is (A) : \(126(5!)^2\)
B₁ , B₂ , B₃ , B₄ , B₅ , B₆ , B₇

Boys can be seated in (7 – 1)! ways = 6! 
Now ways in which no two girls can be seated together is
\(6!\times^7C_5\times5!\)
\(6!\times \frac{7!}{5!2!}\times5!\)
\(=126(5!)^2\)