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Mathematics
In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls may be together ?
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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 ?
BITSAT
BITSAT 2013
A
$4!$
0%
B
$5!$
33%
C
$4! + 5!$
0%
D
$4! \times 5!$
67%
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 $=4 ! \times 5 !$