Corey and Peter on a basketball team. There are 10 players on the team. (a) How many starting lineups (of five players) include both Corey and Peter? (b) How many starting lineups (of five players) include Corey, Peter, or both?
+321 (a) We automatically put Corey and Peter on the team, so we have 3 choices left to make with 8 people to choose from, so the answer is 8C3 = 56.
(b) We want to count these cases seperately. With just Corey we have 8C4 = 70 ways to choose a team and the same goes for Peter. We add 70+70+56 = 196 lineups.
+29 a) A starting lineup of five players that includes Corey and Peter is: Corey, Pater, and three other players. These three must be chosen from the eight that are not Corey and Peter.
Choosing three people out of eight when order doesn't matter makes: \({{8 \cdot 7 \cdot 6} \over {3 \cdot 2 \cdot 1}} = 56\) starting lineups.
a) ANSWER: \(56\) starting lineups.
b) A starting lineup that includes Corey but not Peter is: Corey and four other players who are not Peter. These four must again be chosen from the eight that are not Corey and Peter.
Choosing four people out of eight where order doesn't matter makes: \({{8 \cdot 7 \cdot 6 \cdot 5} \over {4 \cdot 3 \cdot 2 \cdot 1}} = 70\) starting lineups.
A starting lineup that includes Peter but not Corey is: Peter and four other players who are not Corey. There is the same number of these lineups as there are lineups with Corey but not Peter: \(70\) starting lineups.
The number of lineups with both Corey and Peter has already been calculated: \(56\).
So the total number of lineups with Corey, Peter, or both is: \(70 + 70 + 56 = 196\).
b) ANSWER: \(196\) starting lineups.
I hope this is helpful,
Catarina 
