5 boys and 5 girls are sitting in a row randomly. probability that boys and girls sit alternatively

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14.Probability

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$5$ boys and $5$ girls are sitting in a row randomly. The probability that boys and girls sit alternatively is

$5/126$

$1/126$

$4/126$

$6/125$

Solution

(b) Let $n = $ total no. of ways = $10\,!$

$m =$ favourable no. of ways = $2 × 5\,! . \,5\,!$

Since the boys and girls can sit alternately in $5 \,!\ . \,5\,!$ ways if we begin with a boy and similarly they can sit alternately in $5 \,!\ . \,5\,!$ ways if we begin with a girl

Hence, required probability $= \frac{m}{n}= \frac{{2 \times 5!.5!}}{{10!}}$

$= \frac{{2 \times 5!}}{{10 \times 9 \times 8 \times 7 \times 6}} = \frac{1}{{126}}$.

Standard 11

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Solution

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