Equation solution combinatorics

**Greg98** · February 2nd 2011, 07:27 AM

Hello,
the task is to calculate, how many solutions $(x_1, x_2,..., x_n) \in \mathbb{Z}^n$ there 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.

**abhishekkgp** · February 2nd 2011, 09:39 AM

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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