Question 963993
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