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Post Reopened by Mike Earnest combinatorics
Post Closed as "Duplicate" by Parcly Taxel combinatorics

extended Extended stars-and-bars problem(where the upper limit of the variable is bounded)

The problem of counting the solutions $(a_1,a_2,\ldots,a_n)$ with integer $a_i\geq0$ for $i\in\{1,2,\ldots,n\}$ such that $$a_1+a_2+a_3+\ldots+a_n=N$$ can be solved with a stars-and-bars argument. What

What is the solution if one adds the constraint that $a_i\leq r_i$$a_i\leq r_{i}$ for certain integers $r_1,\ldots,r_n$$r_{1},\ldots,r_n$?

e.g. for $n=3$, $N=6$ and $(r_1,r_2,r_3)=(3,3,2)$, the tuple $(a_1,a_2,a_3)=(2,3,1)$ is a solution, but $(2,1,3)$ is not a solution because $a_3=3>2=r_3$$a_{3}=3>2=r_3$.

extended stars-and-bars problem(where the upper limit of the variable is bounded)

The problem of counting the solutions $(a_1,a_2,\ldots,a_n)$ with integer $a_i\geq0$ for $i\in\{1,2,\ldots,n\}$ such that $$a_1+a_2+a_3+\ldots+a_n=N$$ can be solved with a stars-and-bars argument. What is the solution if one adds the constraint that $a_i\leq r_i$ for certain integers $r_1,\ldots,r_n$?

e.g. for $n=3$, $N=6$ and $(r_1,r_2,r_3)=(3,3,2)$, the tuple $(a_1,a_2,a_3)=(2,3,1)$ is a solution, but $(2,1,3)$ is not a solution because $a_3=3>2=r_3$.

Extended stars-and-bars problem(where the upper limit of the variable is bounded)

The problem of counting the solutions $(a_1,a_2,\ldots,a_n)$ with integer $a_i\geq0$ for $i\in\{1,2,\ldots,n\}$ such that $$a_1+a_2+a_3+\ldots+a_n=N$$ can be solved with a stars-and-bars argument.

What is the solution if one adds the constraint that $a_i\leq r_{i}$ for certain integers $r_{1},\ldots,r_n$?

e.g. for $n=3$, $N=6$ and $(r_1,r_2,r_3)=(3,3,2)$, the tuple $(a_1,a_2,a_3)=(2,3,1)$ is a solution, but $(2,1,3)$ is not a solution because $a_{3}=3>2=r_3$.

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joriki
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The problem of counting the solutions $(a_1,a_2,\ldots,a_n)$ with integer $a_i\geq0$ for $i\in\{1,2,\ldots,n\}$ such that $$a_1+a_2+a_3+....a_n=N$$$$a_1+a_2+a_3+\ldots+a_n=N$$ can be solved with a stars-and-bars argument. What is the solution if one adds the constraint that $a_i\leq r_i$ for certain integers $r_1,\ldots,r_n$?

e.g. for $n=3$, $N=6$ and $(r_1,r_2,r_3)=(3,3,2)$, the tuple $(a_1,a_2,a_3)=(2,3,1)$ is a solution, but $(2,1,3)$ is not a solution because $a_3=3>2=r_3$.

The problem of counting the solutions $(a_1,a_2,\ldots,a_n)$ with integer $a_i\geq0$ for $i\in\{1,2,\ldots,n\}$ such that $$a_1+a_2+a_3+....a_n=N$$ can be solved with a stars-and-bars argument. What is the solution if one adds the constraint that $a_i\leq r_i$ for certain integers $r_1,\ldots,r_n$?

e.g. for $n=3$, $N=6$ and $(r_1,r_2,r_3)=(3,3,2)$, the tuple $(a_1,a_2,a_3)=(2,3,1)$ is a solution, but $(2,1,3)$ is not a solution because $a_3=3>2=r_3$.

The problem of counting the solutions $(a_1,a_2,\ldots,a_n)$ with integer $a_i\geq0$ for $i\in\{1,2,\ldots,n\}$ such that $$a_1+a_2+a_3+\ldots+a_n=N$$ can be solved with a stars-and-bars argument. What is the solution if one adds the constraint that $a_i\leq r_i$ for certain integers $r_1,\ldots,r_n$?

e.g. for $n=3$, $N=6$ and $(r_1,r_2,r_3)=(3,3,2)$, the tuple $(a_1,a_2,a_3)=(2,3,1)$ is a solution, but $(2,1,3)$ is not a solution because $a_3=3>2=r_3$.

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Vepir
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Improve formulation
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Marc van Leeuwen
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