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Four couples (husband and wife) decide to form a committee of four members.Find the number of different committees that can be formed in which no couple finds a place is?

My attempt:In one case where one man and three women are chosen our choice for $3$ other committee members becomes $3$ out of $3$ women as we exclude the wife of the first man. Now following the same logic for the next two cases (two men,two women and three men,one woman) and finally adding the other two possibilities where either $4$ men or $4$ women are selected. The final answer came out to be $$2+(4C1 \cdot 3C3)+ (4C2 \cdot 2C2)+(4C3 \cdot 3C3)=24$$ But the answer is $16$. Where did I go wrong?

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  • $\begingroup$ Your answer actually does sum up to be $2+4+6+4=16$. But look at Sakthi's answer for a much simpler calculation of how to do it. $\endgroup$
    – Brent
    Apr 10 '15 at 10:12
  • $\begingroup$ ok. my calculations were wrong. That is so shameful on a website like this. $\endgroup$
    – GrandAlpha
    Apr 10 '15 at 10:58
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This is like choosing only one ball from each of 4 bags where each bag has a red and a blue ball.

Because each couple can have one of the two in the committee (each of the 4 couples have 2 possible choices), by the product rule of counting, $2^4=16$ different committees can be formed.

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