Question
Six boys and six girls sit in a row. What is the probability that the boys and girls sit alternatively
Answer: Option A
:
A
Let n = total number of ways = 12!
and m = favourable numbers of ways = 2×6!.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 = mn=2×6!.6!12!=1462
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:
A
Let n = total number of ways = 12!
and m = favourable numbers of ways = 2×6!.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 = mn=2×6!.6!12!=1462
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