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Q. In how many ways can $5$ boys and $5$ girls be seated at a round table so that no two girls may be together ?

BITSATBITSAT 2013

Solution:

Leaving one seat vacant between two boys, $5$ boys may be seated in $4 !$ ways. Then at remaining $5$ seats, $5$ girls any sit in $5 !$ ways.
Hence the required number $=4 ! \times 5 !$