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# Four couples (husband and wife) decide to form a committee of 4 members. How many ways this can be done so that no couple finds a place.

$\begin{array}{1 1} 10 \\ 12 \\ 14 \\ 16 \end{array}$

The committee can have:
All husbands = $1 \:way$ or All wives = 1 $way$ or
3 husbans and 1 wife (not of the 3 husbands)=$^4C_3\times 1=4\:ways$ or
2 husbands and 2 wifes (not of the two husbands' = $^4C_2\times 1=6 \:ways$ or
1 husband and 3 wives (not the wife of the husband) = $^4C_1\times 1=4\:ways$
$\therefore\:$Total no. of ways the committee can be formed =
$1+1+4+6+4=16\:ways$