There are five boys and three girls who are sitting together to discuss a management problem at a round table. In how many ways can they sit around the table so that no two girls are together?
Solution
The five boys can be arranged on a round table in (5-1)! ways i.e 24
Now there will be five spaces created between two boys.
So three girls can be seated in $$^5C_3$$ ways*3! = 10*6=60
Total number of ways in which five boys and three girls can be seated = 24*60 = 1440
D is the correct answer.
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