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Math Help - Permutations or Combinations? 476

  1. #1
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    Permutations or Combinations? 476

    Page 476
    5 Different colored (black, red, blue, green, and yellow) cards are placed in a row so that the black card is never at either end, how many different arrangements are possible? I don't get why the correct answer is 72.

    Please show steps thank you!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fabxx View Post
    Page 476
    5 Different colored (black, red, blue, green, and yellow) cards are placed in a row so that the black card is never at either end, how many different arrangements are possible? I don't get why the correct answer is 72.

    Please show steps thank you!
    the black card can be in one of 3 positions. 2nd, 3rd, or 4th. for each of those positions, the other 4 cards can be permuted among the other four positions.

    hence, the number of arrangements is 3*4! = 72
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  3. #3
    Kai
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    Hello Fabxx,

    I believe there are other ways to do it, but i did like this...

    First calculate total number of arrangements possible : 5!=120

    Then calculate the number of arrangements for black to be first: 1*4!=24

    And now the number of arrangements for black to be last : 4!*1=24

    So Ur answer would be 120-(24+24)= 72
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