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5 boys and 5 girls sit in a row at random. The probability that the boys and girls sit alternatively is

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\frac{5}{14}

\frac{3}{28}

\frac{1}{11}

\frac{1}{126}

answer is C.

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Detailed Solution

I : B_{1} \times B_{2} \times B_{3} \times B_{4} \times B_{5} \times

$I I : G_{1} \times G_{2} \times G_{3} \times G_{4} \times G_{5} \times$

$∴ n (A) = 2 \times 5! \times 5!, n (S) = 10!$

$P (A) = \frac{n (A)}{n (S)} = \frac{2 \times 5! 5!}{10!} = \frac{1}{126}$

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