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Math Help - permutation/combination question again

  1. #1
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    permutation/combination question again

    5 people stand in line..
    A)how many way are there to arrange these people
    B)if A and B must stand together, how many ways are there to arrange these people
    C)what is the probability A and B stand together

    i think i may have this one figured out, but would like some confirmation that i did it correctly

    A) 5!=120
    B) i grouped A and B together making my sample AB,C,D,E
    -then i did P4/4=4!=24
    -then i considered that AB can be grouped in two ways so did p2/2=2!=2
    -then i multiplied the results 24x2=48 combinations
    C) 48/120

    is this correct?
    i appreciate any input
    thank you
    -ray
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  2. #2
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    Re: permutation/combination question again

    you are right
    an other method for (B)
    AB c d e
    c AB d e
    c d AB e
    c d e AB

    we have 4 x 2 = 8 posibilities and for each (3!) => 8 x 3! = 8 x 6 = 48
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  3. #3
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    Re: permutation/combination question again

    Quote Originally Posted by salim View Post
    you are right
    an other method for (B)
    AB c d e
    c AB d e
    c d AB e
    c d e AB

    we have 4 x 2 = 8 posibilities and for each (3!) => 8 x 3! = 8 x 6 = 48
    thank you
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