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  1. #1
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    How many solutions

    How many solutions does the equation

    x+y+z=100

    have on the set of positive integers?
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by perash View Post
    How many solutions does the equation

    x+y+z=100

    have on the set of positive integers?
    First of all transform the equation to a+b+c = 97.
    Now the solution to this is (97+3-1) choose (3-1) = 99 choose 2
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  3. #3
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    Quote Originally Posted by perash View Post
    How many solutions does the equation

    x+y+z=100

    have on the set of positive integers?
    The equation x_1+x_2+...+x_k = n can be inverstigated in the following way. First construct a row of n 1's: 1 \ 1 \ 1 ... \ 1.
    Now between any space you can place a break \bold{b} and that will give you a possible (x_1,...,x_k) pair.
    For example, to illustrate, n=5, k=3, we can break it as 1 \ 1 \ \bold{b} \ 1 \ \bold{b} \ 1 \ 1. That gives us x_1=2, x_2=1,x_3=2. Thus, given n as an integer we place k-1 breaks in between the n-1 breaks. There are a total of {{n-1}\choose{k-1}} ways of doing this.

    @Isomorphism: The proof of the combinatorial formula you used rests on what I did.
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by ThePerfectHacker View Post
    @Isomorphism: The proof of the combinatorial formula you used rests on what I did.
    I know that
    But I was too lazy to write that argument out. Isn't there another way too look at it-The coefficient of polynomial way?
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  5. #5
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    Hello, perash!

    How many solutions does the equation: . x+y+z=100
    have on the set of positive integers?
    Here's a "visual" explanation of Isomorphism's solution . . .


    We have a 100-inch "yardstick", marked in inches.

    We will cut it into three pieces by making two cuts.

    There are 99 inch-marks and we will chose two at which to cut.

    Answer: . {99\choose2} \:=\:4851 ways.

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