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(B) Total number of ways to arrange $10$ people in a row is $n = 10!$.For boys and girls to sit alternatively,they must occupy positions like $(B, G,

Vedclass · Accepted Answer

$1/126$

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(B) Total number of ways to arrange $10$ people in a row is $n = 10!$.For boys and girls to sit alternatively,they must occupy positions like $(B, G, B, G, B, G, B, G, B, G)$ or $(G, B, G, B, G, B, G, B, G, B)$.Case $1$: Starting with a boy,the number of arrangements is $5! 	imes 5!$.Case $2$: Starting with a girl,the number of arrangements is $5! 	imes 5!$.Total favourable ways $m = 2 	imes (5! 	imes 5!)$.Required probability $P = \frac{m}{n} = \frac{2 	imes 5! 	imes 5!}{10!}$.$P = \frac{2 	imes 120 	imes 120}{3628800} = \frac{2 	imes 120}{10 	imes 9 	imes 8 	imes 7 	imes 6} = \frac{2}{30240} 	imes 120 = \frac{2}{252} = \frac{1}{126}$.

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

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