A certain baseball team has 19 players. Only nine can be on the field at a time.
ID: 3047495 • Letter: A
Question
A certain baseball team has 19 players. Only nine can be on the field at a time. Each of the nine players on the field has a distinct field position: pitcher, catcher, first baseman, second baseman, third baseman, short stop, left field, right field, or center field. Assume for the moment that every player is qualified to play every position.
a) How may ways are there to select nine of the 19 players to be on the field (without regard to which position each player will have)?
b) How many ways are there to assign field positions to a given set of nine players?
c) How many ways are there to fill the nine field positions from among the 19 players (taking account of the position each player will have)?
d) How many ways are there to fill either the pitcher or catcher field position (but not both) from among the 19 players (leaving the other field positions empty)?
Now suppose that not every player can play in every position. The outfielders (left field, center field, right field) can play any outfield position, the infielders (1st base, 2nd base, 3rd base, short stop) can play any infield position, the pitchers can only pitch, and the catchers can only catch. Suppose a certain team has 20 players, of whom 3 are catchers, 4 are outfielders, 6 are infielders, and 7 are pitchers.
e) How many ways can the team assign field positions to 9 of the 20 players, putting each of the 9 selected players in a position he can play, and ensuring that all 9 field positions are filled?
Explanation / Answer
a)
C(19,9) = 92378 ways
b)
9! ways = 362880 ways
c)
P(19,9) = 33522128640 ways
d)
There are 19*18 = 342 ways
PS : We are directed to solve only the first question when multiple questions are posted or the first 4 parts when multiple parts of the same question are posted.
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