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6 boys and 6 girls sit randomly in a row, what is thae probability that boys and girls sit alternately?

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1 Answer

Given: 6 boys and 6 girls.
$\therefore\:$ There are 12 persons.
All of them can be seated in $(12)!$ ways.
$i.e.,n(S)=(12)!$ where $S$ is sample space.
$E=$ Event of the girls and boys to be seated alternatively.
6 girls are seated in $6!$ ways. $G_1, G_2,G_3,G_4,G_5,G_6$
There are $7$ alternate places and $6$ Boys.
$\therefore$ $6$ boys can be seated in $7$ places in $^7P_6$ ways.
$\therefore\: n(E)=6!.^7P_6=\large\frac{6!.7!}{(7-6)!}$$=6!.7!$
$\therefore$ The required probability=$\large\frac{n(E)}{n(S)}=\frac{6!.7!}{(12)!}$
answered Feb 15, 2014 by rvidyagovindarajan_1

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