A) \[\frac{1}{462}\]
B) \[\frac{1}{924}\]
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: A
Solution :
Let \[n=\] total number of ways \[=12\,!\] and \[m=\] favourable number 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 \[=\frac{m}{n}=\frac{2\times 6\,!\,\,.\,\,6\,!}{12\,!}=\frac{1}{462}\].
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