Question 963993: Here's the question:
A baseball team has 15 members. Five of the players are pitchers, and the remaining 10 members can play any position. How many different teams of 9 players can be formed?
I know this will be a Combination because order doesn't matter, but I don't know where the pitchers and normal players come in. How many pitchers are required to be on the team?
I know that if the pitchers and normal players don't matter, the answer is simply 15C9. But do the pitchers matter?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 9 players are on the field at one time.
there is 1 pitcher, 1 catcher, 3 outfielders and 4 infielders, for a total of 9.
i believe you have to make that assumption.
so each team will have 1 pitcher and 8 other members on the team.
order doesn't matter.
the number of different possible 9 member teams should be:
(5c1) * (9c8) = 45.
look at this in a much smaller type problem and you can see how it works.
assume 2 pitchers and 4 other players.
assume 3 man teams of 1 pitcher and 2 other players.
number of possible teams should be 2c1 * 4c2 = 12
let the pitchers be a and b
let the other players be 1 and 2 and 3 and 4.
the possible 3 man teams are:
a12, a13, a14, a23, a24, a34
b12, b13, b14, b23, b24, b34
each team has 1 pitcher and 2 other players.
each team is different.
the formula is shown to work for the smaller problem.
good chance it will work for the bigger problem.
5c1 * 9c8 = 45
5c1 = combination formula of 5! / (1! * 4!) = 5
9c8 = combination formula of 9! / (8! * 1!) = 9
5 * 9 = 45
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