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

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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 =

$1+1+4+6+4=16\:ways$

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