A committee of $5$ members is to be chosen seven men and five women. Find the number of ways this committee can be formed if the man Aaron or the woman Beatrice or both of them must be in the committee.
My solution
$${12\choose 5} - {7\choose5} - {5\choose5} = 770$$
My answers is incorrect, it should be $540$. I realised that I only subtracted the numbers of ways for no men and no women from the total number of ways. How can I get the answer from the total numbers of ways?
On
The complement approach shown by @SalmonFish is the simplest, but here's an alternative (inclusion-exclusion) approach. The number of committees that include Aaron is $\binom{12-1}{5-1}=\binom{11}{4}$. The number of committees that include Beatrice is also $\binom{11}{4}$. If you just add these two, you have overcounted because each committee that includes both is counted twice. The number of committees that include both Aaron and Beatrice is $\binom{12-2}{5-2}=\binom{10}{3}$. Putting it all together, the number of committees that include either Aaron or Beatrice (or both) is $$\binom{11}{4} + \binom{11}{4} - \binom{10}{3} = 330 + 330 - 120 = 540.$$
On
Both @SalmonFish and @RobPratt have supplied nice solutions. Here is yet another approach:
There are two possibilities:
Exactly one of Aaron and Beatrice is selected: We can choose one person from among Aaron and Beatrice and $5 - 1 = 4$ of the other $5 + 7 - 2 = 10$ people in $$\binom{2}{1}\binom{10}{4}$$ ways.
Both Aaron and Beatrice are selected: We can choose both Aaron and Beatrice and $5 - 2 = 3$ of the other $10$ people in $$\binom{2}{2}\binom{10}{3}$$ ways.
Total: Since these cases are mutually exclusive and exhaustive, the number of ways we can select a five-person committee which contains Aaron, Beatrice, or both of them is $$\binom{2}{1}\binom{10}{4} + \binom{2}{2}\binom{10}{3} = 540$$
Let's use complement rule: all selections without restriction minus neither Aaron nor Beatrice in the committee at the same time. When we subtract them, we have $10$ people to select $5$ people among them.
Then $$\binom{12}{5} - \binom{10}{5} = 540.$$