$\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.

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