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Thread: 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 $\displaystyle \: = \frac {8!}{2!}\: =\: 20160.$

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

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

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

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

    What next?

    Thanks.

    ILoveMaths07.
    Last edited by ILoveMaths07; Sep 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 $\displaystyle C$ and $\displaystyle A$ are together, but $\displaystyle N$ and $\displaystyle D$ are not.
    First, find the number of permuations in which $\displaystyle C$ and $\displaystyle A$ are together.

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

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

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



    How many have $\displaystyle C$ and $\displaystyle A$ together and $\displaystyle N$ and $\displaystyle D$ together?

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

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

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

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


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

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