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In a building with 10 couples of man and woman, how many different committees of 6 people we can make such that it will have at least one man and at least one woman and no spouses?

My attempt with complement:

Choose 6 people out of 20: $\binom {20} 6$.

No women: $\binom {10} 6$, no men $\binom {10} 6$.

Only spouses: $\binom {10} 3$.

Total: $\binom {20} 6-(2\binom {10} 6+\binom {10} 3)$ committees.

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  • $\begingroup$ $6$ couples or $6$ people? $\endgroup$ Commented Mar 8, 2015 at 12:38
  • $\begingroup$ 6 people in a committee. @barakmanos $\endgroup$
    – GinKin
    Commented Mar 8, 2015 at 12:38
  • $\begingroup$ I think that the "only spouses" part is wrong, because you are not excluding cases where some are spouses and some are not. $\endgroup$ Commented Mar 8, 2015 at 12:40
  • $\begingroup$ @barakmanos how can this be done? $\endgroup$
    – GinKin
    Commented Mar 8, 2015 at 12:47

1 Answer 1

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Choose $6$ different couples to provide a person for the committee ($10\choose 6$ ways).

Choose $1$ person from each of the $6$ couples, but exclude the choice of all women and the choice of all men ($2^6-2$ ways).

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  • $\begingroup$ The total is: $\binom {10} 6(2^6-2)$ ? $\endgroup$
    – GinKin
    Commented Mar 8, 2015 at 12:46
  • $\begingroup$ Yes, I think so. $\endgroup$
    – paw88789
    Commented Mar 8, 2015 at 12:47

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