Question

Six boys and six girls sit in a row at random. The probability that the boys and girls sit alternatively is

• A. $\frac{1}{462}$
• B. $\frac{1}{231}$
• C. $\frac{1}{346}$
• D. $\frac{1}{158}$

Answer 1

The correct option is A $\frac{1}{462}$

$n\left(s\right)=12!$
$n\left(E\right)=2(6!6!)$
$P\left(E\right)=\frac{1}{462}$