Let $v=(v_1,...,v_k)\in\mathbb{Z}_{>0}^k$. Denote by $l(a)$ the sum of the entries of an integer vector $a$ and consider on $\mathbb{Z}_{>0}^k$ the order $\leq$ defined by coordinatewise comparison, i.e. $a\leq b\Leftrightarrow \forall j\in [k]: a_j\leq b_j$. I would like to count for a fixed $m\leq l(v)$ the number of vectors $w\in\mathbb{Z}_{>0}^k$ s.t. $l(w)=m$ and $w\leq v$. I need a combinatorial proof/way of calculating this. I have tried a couple of things now but nothing really works out in a nice enough fashion. This is no homework or something like that. I appreciate any help and input. Thanks in advance!
This is equivalent to counting the number of number-compositions of $m$ with $k$ summands where there is an upper bound $v_i$ on the i-th summand. If that helps.
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1If nonnegative integer vectors instead of positive integers, your question would be this question exactly. It should be clear how to translate your problem to the linked one. – Mike Earnest Oct 08 '21 at 14:21
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Thank you very much, that confirms my apprehension that there is no closed form solution. – Mo145 Oct 09 '21 at 10:46