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Math Help - A counting problem

  1. #1
    MHF Contributor alexmahone's Avatar
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    A counting problem

    In how many ways can we list the digits {1,1,2,2,3,4,5} so that two identical digits are not in consecutive positions?

    My solution:

    No. of ways in which the 1's are in consecutive positions = 6!/2! = 360
    No. of ways in which the 2's are in consecutive positions = 6!/2! = 360
    No. of ways in which both 1's and 2's are in consecutive positions = 5! = 120
    No. of ways in which either the 1's or 2's are in consecutive positions = 360+360-120=600
    Total number of permutations = 7!/(2!2!) = 1260
    No. of ways in which two identical digits are not in consecutive positions = 1260-600=660

    Could someone please check if my solution is correct? Thanks.
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  2. #2
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    Re: A counting problem

    Quote Originally Posted by alexmahone View Post
    In how many ways can we list the digits {1,1,2,2,3,4,5} so that two identical digits are not in consecutive positions?
    No. of ways in which the 1's are in consecutive positions = 6!/2! = 360
    No. of ways in which the 2's are in consecutive positions = 6!/2! = 360
    No. of ways in which both 1's and 2's are in consecutive positions = 5! = 120
    No. of ways in which either the 1's or 2's are in consecutive positions = 360+360-120=600
    Total number of permutations = 7!/(2!2!) = 1260
    No. of ways in which two identical digits are not in consecutive positions = 1260-600=660
    That is correct.
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