$6$ boys and $5$ girls are to sit in a row at random.The probability that boys and girls sit alternately is

\frac{5! 5!}{11!}

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\frac{6! 6!}{11!}

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\frac{6! 5!}{11!}

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1 - \frac{6! 5!}{11!}

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Solution

The correct option is A $\frac{6! 5!}{11!}$
Arrangement in which boys and girls sit alternately: $B_{1} G_{1} B_{2} G_{2} B_{3} G_{3} B_{4} G_{4} B_{5} G_{5} B_{6}$
We could arrange boys in $6!$ ways and girls in $5!$ ways
Probability $= \frac{6! 5!}{11!}$

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