# How many different committees with at least one man and woman and no spouses

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.

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

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).
• The total is: $\binom {10} 6(2^6-2)$ ? Commented Mar 8, 2015 at 12:46