A woman has $10$ friends, of whom $5$ are women and $5$ are men. In how many ways can she invite three or more to a party? In how many ways can she invite three or more to a party if she wants the same number of men as women (including herself)?
My attempt so far:
I am unsure on how to answer if there are $3$ or more people.
If she chooses $3$ men and two women = $6$ people including herself $\dbinom{5}{3}=10$ $\dbinom{5}{2}=10$
The answer would be $20$ different combinations?
Hint: In this case, the gender of the friends she invites does not matter. Therefore, the woman invites three of her ten friends or four of her ten friends or five of her ten friends or ...
Hint: Since she invites at least three friends, she must two of the five men and one of the five women or three of the five men and two of the five women or four of the five men and three of the five women or all five men and four of the five women.
The Multiplication Principle states that when one task can be performed in $m$ ways and a second task can be performed independently of the first in $n$ ways, then there are $mn$ ways of performing both tasks.
The number of ways the woman can select which three men to invite and which two women to invite are independent. Therefore, the number of ways of inviting three of the five men and two of the five women to the party is $$\binom{5}{3}\binom{5}{2}$$
The Addition Principle states that if one task can be done in $m$ ways and a second task that cannot be done at the same time as the first task can be done in $n$ ways, there are $m + n$ ways of performing one of these tasks.
For instance, the woman can invite three of her ten friends in $\binom{10}{3}$ ways. She can invite four of her ten friends in $\binom{10}{4}$ ways. Therefore, the number of ways the woman can invite three or four of her ten friends is $$\binom{10}{3} + \binom{10}{4}$$
The word and is an indication that you should multiply, while the word or is an indication that you should add.