InterviewSolution
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InterviewSolution
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Find the number of ways of dividing 6 couples in 3 groups if each group has exactly one couple and each group has 2 males and 2 females. |
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Answer» There are 6 couples `(A_(1)B_(1)),(A_(2)B_(2)),(A_(3)B_(3)),(A_(4)B_(4)),(A_(5)B_(5)) " and " (A_(6)B_(6)), " where " A_(i)` is male and `B_(i)` is female. We have to divide these couples (12 persons) in three groups P, Q and R, such that each containing exactly one couple having two males and two females. First select three couples for three groups P, Q, R. Number of ways are `""^(6)C_(3)`. Now from the remaining three couples three males can enter in groups P, Q and R in 3! ways. The remaining three females cannot enter the group in which their spouses have entered. So, for females we have derangement `D(3)=3!(1-(1)/(1!)+(1)/(2!)-(1)/(3!))=2` Therefore, total number of required ways `= ""^(6)C_(3)xx3!xx2` =240 |
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