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Math Help - Equation solution combinatorics

  1. #1
    Junior Member Greg98's Avatar
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    Equation solution combinatorics

    Hello,
    the task is to calculate, how many solutions (x_1, x_2,..., x_n) \in \mathbb{Z}^nthere is to equation x_1+x_2+...+x_n=k. Every solution must be x_i \geq i for all i=1,2,...,n.

    I got something like this:
    y_i=x_i-i \geq 0
    (x_1-1)+(x_2-2)+...+(x_n-n)=k-\frac{n(n+1)}{2}
    y_1+y_2+...+y_n=k-\frac{n(n+1)}{2}
    \Rightarrow \left(\begin{array}{cc}\frac{n(n+1)}{2}+k-1\\k\end{array}\right)

    Is that right or Am I missing something? Any help is appreciated. Thank you.
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  2. #2
    Senior Member abhishekkgp's Avatar
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    my answer would be \Rightarrow \left(\begin{array}{cc}k-\frac{n(n+1)}{2}+n-1\\n-1\end{array}\right)

    this is because the number of integral solutions to the equation b_1+b_2+ \ldots + b_n=q, where b_i \geq 0 \text{ } \forall \text{ } i is given by \left(\begin{array}{cc}q+n-1\\n-1\end{array}\right)
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