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Math Help - Should I use permutation or combination in here?

  1. #1
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    Should I use permutation or combination in here?

    A certain baseball team has 24 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.


    How may ways are there to select nine of the 24 players to be on the field (without regard to which position each player will have)?

    My reasoning is: 24x23x22x21x20x19x18x17x16 =474467051520

    I don't get why this is wrong though. Any help would be appreciated!
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    I think it's 9*(24 choose 9)
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  3. #3
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Felipe View Post
    A certain baseball team has 24 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.


    How may ways are there to select nine of the 24 players to be on the field (without regard to which position each player will have)?

    My reasoning is: 24x23x22x21x20x19x18x17x16 =474467051520

    I don't get why this is wrong though. Any help would be appreciated!
    You would have the right answer if position mattered, but since it says "without regard to which position each player will have" you must divide your answer by 9! to make it right. This is the same as doing \displaystyle \binom{24}{9}. That is, \displaystyle \binom{24}{9}=\frac{24!}{9!(24-9)!}=\frac{24\cdot23\cdot22\cdot21\cdot20\cdot19\c  dot18\cdot17\cdot16}{9!}=1307504. (Edited a bit for clarity.)
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