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Aman

asked 2017-01-30 14:33:29

There are $4$ boys and $4$ girls. They sit in a row randomly. What is the chance that all the girls do not sit together.

jiju

answered 2017-01-31 20:01:44

Without any restrictions, number of possible arrangements $=8!$

Group the $4$ girls and consider as a single girl. then there are $5$ persons and they can be arranged in $5!$ ways. The $4$ girls can be arranged among themselves in $4!$ ways.

Therefore, number of ways in which all the $4$ girls will always sit together

$=5!×4!$

Hence, number of ways in which all the $4$ girls do not sit together

$8!-5!×4!$

Required probability

$=\dfrac{8!-5!×4!}{8!}=\dfrac{13}{14}$

Group the $4$ girls and consider as a single girl. then there are $5$ persons and they can be arranged in $5!$ ways. The $4$ girls can be arranged among themselves in $4!$ ways.

Therefore, number of ways in which all the $4$ girls will always sit together

$=5!×4!$

Hence, number of ways in which all the $4$ girls do not sit together

$8!-5!×4!$

Required probability

$=\dfrac{8!-5!×4!}{8!}=\dfrac{13}{14}$

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