Question

Six boys and six girls are to sit around a roundtable. The probability that boys and girls sit alternatively

• A. $\frac{5!5!}{11!}$
• B. $\frac{5!6!}{11!}$
• C. $\frac{6!6!}{12!}$
• D. $\frac{5!6!}{12!}$

Answer 1

Answer: (B)

$\frac{5!6!}{11!}$

Only one arrangement possible: bgbgbgbgbgbg
First arrange girls and then place $6$ boys between $6$ girls
The ways of doing this $=(6-1)!\times 6!$
The total arrangements in circle $=(12-1)!$
The required probability $=\frac{(6-1)!\ast 6!}{11!}$