Six boys and six girls sit in a row. What is | vedclass

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Question

(a) Let $n = $ total number of ways $ = 12\,!$

and $m = $ favourable number of ways $ = 2 	imes 6\,!\,.\,6\,!$

Since the boys and girls can sit

Vedclass · Accepted Answer

$\frac{1}{{462}}$

Vedclass · Answer

(a) Let $n = $ total number of ways $ = 12\,!$

and $m = $ favourable number of ways $ = 2 	imes 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 $ = \frac{m}{n} = \frac{{2 	imes 6\,!\,\,.\,\,6\,!}}{{12\,!}} = \frac{1}{{462}}$.

Six boys and six girls sit in a row. What is the probability that the boys and girls sit alternatively

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