Four couples (husband and wife) decided. Find the number of different committees that can be formed in which no couple finds a place.

The correct option is C

$16$

Explanation for the correct option:

Find the number of different committees :

There are four ways to form a committee of four-member

All are gentlemen, $3$husband, $1$wife, $2$husband, $2$wife, and all are wives.

The number of committees of $4$ gentlemen $\Rightarrow {}_{}{}_{}{}^{4}C_4{}_{}{}^{}=1$

The number of committees of $3$gentlemen, $1$wife $\Rightarrow {}_{}{}_{}{}^{4}C_3{}_{}{}^{}\times {}_{}{}_{}{}^{1}C_1{}_{}{}^{}=4$

The number of committees of $2$gentlemen, $2$wife$\Rightarrow {}_{}{}_{}{}^{4}C_2{}_{}{}^{}\times {}_{}{}_{}{}^{2}C_2{}_{}{}^{}=6$

The number of committees of $4$wives $\Rightarrow {}_{}{}_{}{}^{4}C_4{}_{}{}^{}=1$

The required number of committees $\Rightarrow 1+4+6+4+1=16$
The number of different committees that can be formed in which no couple finds a place is $16.$

Hence, the correct option is (C).