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Math Help - Permutation of the letters ABCDEFGH

  1. #1
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    Arrow Permutation of the letters ABCDEFGH

    Can anyone please check my below answers!

    Q.1 How many permutations of the letters ABCDEFGH contain

    a) the strings CAB and BED?

    myans.) 5P5 = 5! = 120 ways

    b) the strings BCA and ABF?

    myans.) permutation not possible since B cannot be followed by both C and F at the same time.

    Thanks for any help!
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  2. #2
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    Quote Originally Posted by robocop_911 View Post
    Can anyone please check my below answers!

    Q.1 How many permutations of the letters ABCDEFGH contain

    a) the strings CAB and BED?

    myans.) 5P5 = 5! = 120 ways

    b) the strings BCA and ABF?

    myans.) permutation not possible since B cannot be followed by both C and F at the same time.

    Thanks for any help!
    Well, the letters CAB can appear in 6 positions...

    CABxxxxx, xCABxxxx, xxCABxxx, xxxCABxx, xxxxCABx and xxxxxCAB

    BED cannot appear in the last 2 positions, so there are only 4 positions that permit CAB and BED.

    Of these, the remaining letters have 3! permutations each...

    So there are 4*6 = 24 permutations allowing the combination CABED.
    Last edited by sean.1986; May 30th 2008 at 09:18 PM.
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  3. #3
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    Exclamation

    Quote Originally Posted by robocop_911 View Post
    Can anyone please check my below answers!

    Q.1 How many permutations of the letters ABCDEFGH contain

    a) the strings CAB and BED?

    myans.) 5P5 = 5! = 120 ways

    b) the strings BCA and ABF?

    myans.) permutation not possible since B cannot be followed by both C and F at the same time.

    Thanks for any help!
    I take my word back!
    Last edited by robocop_911; June 1st 2008 at 04:32 PM.
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  4. #4
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    Quote Originally Posted by sean.1986 View Post
    Well, the letters CAB can appear in 6 positions...

    CABxxxxx, xCABxxxx, xxCABxxx, xxxCABxx, xxxxCABx and xxxxxCAB

    BED cannot appear in the last 2 positions, so there are only 4 positions that permit CAB and BED.

    Of these, the remaining letters have 3! permutations each...

    So there are 4*6 = 24 permutations allowing the combination CABED.


    Can anyone please verify whether whatever "sean1986" has said is Correct or Not! Thanks!
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  5. #5
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    Hello, robocop_911!

    Q.1 How many permutations of the letters ABCDEFGH contain

    a) the strings CAB and BED?

    my answer: . _5P_5 \:= \:5! \:= \:120 ways . . . . not quite

    The permutation must contain \boxed{CABED}.

    Then we have four objects to permute: . \boxed{CABED}\,,\:\boxed{F}\,,\:\boxed{G}\,,\:\box  ed{H}

    And there are: . _4P_4 \:=\:4! \:=\: 24 ways.




    b) the strings BCA and ABF?

    my answer: .The permutation not possible
    since B cannot be followed by both C and F at the same time.

    I agree!

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