I know the answer for the cases where I can have all possible values which will be given by $(n+k-1)C(k-1)$ but I don't know how to reduce it to the given constraints.
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Why not try stars and bars? You have n stars and have to use 2 bars to split them. The first can cover a maximum of n/3 stars and the second can separate another n/3 stars or less. – For the love of maths Dec 25 '19 at 11:05
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@MohammadZuhairKhan This problem has the added constraint that $x \leq y \leq z$. – N. F. Taussig Dec 25 '19 at 11:06
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@N.F.Taussig I added to my comment. – For the love of maths Dec 25 '19 at 11:08
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You're looking for the restricted partition number $p_3(n)$. This is the number of ways to divide $n$ into three integral pieces without regard to order. For every such arrangement, you would assign $x$ to a minimal value of that triple, $z$ to a maximal value, and $y$ to the remaining value.
Mathworld helpfully notes at (66) that $p_3(n)=\left[\frac{n^2}{12}\right]$, where the square bracket indicates rounding to the nearest integer.