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Math Help - combinatoric thing

  1. #1
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    combinatoric thing

    question.

    each morning Mr. Hadi puts on his socks and shoes, and laces up his shoes before going to school. He must put on his left sock before he can put on his left shoe, and the same for his right sock and right shoe, and he must also put each shoe on before he can lace it up. In how many different ways can he do these six things?
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  2. #2
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    S is sock , E is shoe, and L is lace.

    [LaTeX ERROR: Convert failed] is the most obvious method. this can be done is 8 ways, by swapping order of left and right.

    or it could be done like this.

    [LaTeX ERROR: Convert failed] and this can be done in two ways.

    or like this.
    [LaTeX ERROR: Convert failed] and this can we done in a further 4 ways.

    giving a total of 8 + 2 + 4 = 14 ways (i hope)

    bobak.

    Edit: I missed a few things, ignore this.
    Last edited by bobak; April 6th 2008 at 04:27 PM.
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  3. #3
    Junior Member roy_zhang's Avatar
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    Let L1 denote put on his left sock, L2 denote put on left shoe and L3 denote lace up left shoe. R1, R2 and R3 are defined similarly. I come up the following diagram. each path through the tree nodes will represent one possibility, so it has 20 different ways to these six things. Anyone has an "algebraic" way (using addition/multiplication principles, combinations, permutations etc.) to solve this, please share with us.


    Roy
    Attached Thumbnails Attached Thumbnails combinatoric thing-tree.jpg  
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  4. #4
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    sock(left,right),shoe(left,right),lace(left, right),therefore there exist 2+2+2=6 task need to done。
    let a=>sock ,b=>shoe ,c=>lace

    therefore the case seem like this a1b1c1 , a2b2c2
    _ _ _ _ _ _
    let the above space represent the flow of the task。
    firstly, put a1,b1,c1, into the task flow。
    therefore there got 3 out of 6 place to put it in = C(6,3)
    but we need to make sure the flow of a1->b1->c1
    there only have 1 method to arrange a->b->c that is a1b1c1
    thus C(6,3)*1
    if do not need to consider a1->b1->c1
    this will make the number of arrangement become C(6,3)*(3*2*1)
    in our case, should be C(6,3)*1=C(6,3)=20
    -------------------------------------------
    next,the empty task flow remain 3 and simply put remaining a2,b2,c2,
    inside the empty space. However, we need to consider a2->b2->c2
    thus only got 1 arrangement.

    to visualize it.
    a1 _ _ b1 _ c1 -> a1 a2 b2 b1 c2 c1
    _ a1 b1 c1 _ _ - > a2 a1 b1 c1 b2 c2
    a1 _ _ _ b1 c1 -> a1 a2 b2 c2 b1 c1

    --------------------------------
    therefore the answer is [C(6,3)*1]*1 = C(6,3) = 20
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