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Math Help - Permutations

  1. #1
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    Permutations

    Calculate the number of permutations of the word CALENDAR in which C and A are together but N and D are not.

    This is what I did:

    No. of possible permutations \: = \frac {8!}{2!}\: =\: 20160.

    No. of permutations in which C and A are together \: =\: 5! * \frac {3!}{2!}\: =\: 360.

    No. of permutations in which C and A are not together  = 20160 - 360 = 19800.

    No. of permutations in which N and D are together \: = \frac {6!}{2!} * 2! = 720.

    No. of permutations in which N and D are not together  = 20160 - 720 = 19440.

    What next?

    Thanks.

    ILoveMaths07.
    Last edited by ILoveMaths07; September 20th 2008 at 08:02 AM.
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  2. #2
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    Hello, ILoveMaths07!

    Calculate the number of permutations of the word CALENDAR
    in which C and A are together, but N and D are not.
    First, find the number of permuations in which C and A are together.

    Duct-tape C and A together.
    They could be: \boxed{CA} or \boxed{AC} . . . 2 ways.

    We have 7 "letters" to arrange: . \{A,D,E,L,N,R, \boxed{CA}\, \}
    . . They can be arranged in: . {\color{blue}7!} ways.

    Hence, there are: . 2 \times 7! \:=\:10,080 ways to have C and A together.



    How many have C and A together and N and D together?

    Duct-tape C and A together. There are 2 ways: . \boxed{CA} or \boxed{AC}

    Duct-tape N and D together. There are 2 ways: . \boxed{ND} or \boxed{DN}

    We have six "letters" to arrange: . \{A,E,L,R,\boxed{CA},\,\boxed{ND}\, \}
    . . They can be arranged in: {\color{blue}6!} ways.

    Hence, there are: . 2\times 2 \times 6! \:=\:2880 ways to have C,A and N,D together.


    Therefore, there are: . 10,080 - 2,880 \;=\;\boxed{7,200} ways
    . . in which C and A are together, but N and D are not.

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