Question

Six boys and six girls sit in a row. What is the probability that the boys and girls sit alternatively

• A. $\frac{1}{462}$
• B. $\frac{1}{924}$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

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

Let n = total number of ways = 12!
and m = favourable numbers of ways = $2\times 6!.6!$
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
Hence required probability = $\frac{m}{n}=\frac{2\times 6!.6!}{12!}=\frac{1}{462}$