Question

Six boys and girls sit in a row randomly. Find the probability that the boys and girls sit alternately.

• A. $p=\frac{1}{311040}$
• B. $p=\frac{1}{924}$
• C. $p=\frac{1}{462}$
• D. $p=\frac{1}{231}$

Answer 1

Answer: (C)

$p=\frac{1}{462}$

Total number of boys and girls $=6+6=12$
Total cases $n=12!$
Since the boys and girls can sit alternately in 6! 6! ways if we begin with a boy and similarly they can sit alternately in 6! 6! ways if we begin with a girl.
$m=2\times 6!6!$
Hence $p=\frac{m}{n}=\frac{2\times 6!6!}{12!}$
$p=\frac{1}{462}$