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    Joined: Feb 2013
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    Originally Posted by arlen1
    AFAIK, it is 'number of ways to place 4 identical balls into 3 containers' = 'throw 2 sticks between 4 balls, on the outside allowed' = C(6,2) = (6*5)/(2*1) = 15.

    (3 containers, and not 4, because the number < 1000 and so has <= 3 digits.)
    Ah of course. For some reason I misread it as a 4 digit number.

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    We don't know the age and background of the OP's daughter, but it looks like the question is designed for someone with no particular background in combinatorics (counting techniques). Instead, the approach is to just try and list all the numbers systematically, e.g. in increasing order. (The problem said to actually list them, not just count them.) It's most educational if the student just tries to figure it out by themselves without being taught a method.

    In other words there's not supposed to be an already-learned method, and I was trying to make this point in an obscure way by giving a (counting) method.

    If you want a counting method, a classic one is given in the mathforum.org link above. Another classic counting method is the "generating function" method I gave above. Briefly, you want to count solutions to
    a+b+c=4 (a,b,c non-negative integers)
    Now 1/(1-x)=1+x+x^2+x^3+x^4+x^5+...
    Distribute
    (1+x+x^2+x^3+x^4+x^5+...)(1+x+x^2+x^3+x^4+x^5+...)(1+x+x^2+x^3+x^4+x^5+...)
    and collect like terms.
    The fifteen x^4 terms correspond to all the products (x^a)(x^b)(x^c)=x^{a+b+c} equalling x^4.

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    Originally Posted by 22B
    We don't know the age and background of the OP's daughter, but it looks like the question is designed for someone with no particular background in combinatorics (counting techniques). Instead, the approach is to just try and list all the numbers systematically, e.g. in increasing order. (The problem said to actually list them, not just count them.) It's most educational if the student just tries to figure it out by themselves without being taught a method.

    In other words there's not supposed to be an already-learned method, and I was trying to make this point in an obscure way by giving a (counting) method.

    If you want a counting method, a classic one is given in the mathforum.org link above. Another classic counting method is the "generating function" method I gave above. Briefly, you want to count solutions to
    a+b+c=4 (a,b,c non-negative integers)
    Now 1/(1-x)=1+x+x^2+x^3+x^4+x^5+...
    Distribute
    (1+x+x^2+x^3+x^4+x^5+...)(1+x+x^2+x^3+x^4+x^5+...)(1+x+x^2+x^3+x^4+x^5+...)
    and collect like terms.
    The fifteen x^4 terms correspond to all the products (x^a)(x^b)(x^c)=x^{a+b+c} equalling x^4.

    A lovely, succinct explanation. Thank you!


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