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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

$\begin{array}{1 1}(A)\;10\\(B)\;12\\(C)\;14\\(D)\;16\end{array}$

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• $C(n,r)=\large\frac{n!}{r!(n-r)!}$
The number of committees of 4 gentlemen =$4C_4$
$\Rightarrow 1$
The number of committees of 3 gentlemen,1 wife =$4C_3\times 1C_1$
(After selecting 3 gentlemen only 1 wife is left who can be included)
The number of committees of 2 gentlemen 2 wives =$4C_2\times 2C_2$
The number of committees of 1 gentlemen 3 wives =$4C_1\times 3C_3$
The number of committees of 4 wives =1
$\therefore$ The required number of committees =1+4+6+4+1
$\Rightarrow 16$
Hence (D) is the correct answer.