Complement counting

#complement counting#inclusion-exclusion#committee

Statement

From $7$ men and $5$ women, how many committees of $4$ contain at least one man?

Hint

Compute total minus all-women committees.

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Total. $\binom{12}{4}=495$ .

All women. $\binom{5}{4}=5$ .

At least one man. $495-5=490$ .

Conclusion. There are $490$ committees with at least one man.