Four couples (husband and wife) decide to form a committee of four members. The number of different committees that can be formed in which no couple finds a place is

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Solution

The correct option is D 16
Four couples are as follows:
$H_{1} W_{1}, H_{2} W_{2}, H_{3} W_{3}, H_{4} W_{4}$

No couple can find a place when only one member of each group is selected.

Hence no. of ways of doing this = $(\begin{matrix} 44 \end{matrix}) \times (\begin{matrix} 21 \end{matrix}) \times (\begin{matrix} 21 \end{matrix}) \times (\begin{matrix} 21 \end{matrix}) \times (\begin{matrix} 21 \end{matrix}) = 16$

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Four couples (husband and wife) decide to form a committee of four members. The number of different committees that can be formed in which no couple finds a place is

The correct option is D 16Four couples are as follows:H1W1,H2W2,H3W3,H4W4No couple can find a place when only one member of each group is selected.Hence no. of ways of doing this =(44)×(21)×(21)×(21)×(21)=16