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Let $N \in \mathbb{Z}_{\geq 0}$ and let $\alpha = (a_1,...,a_n) \in \mathbb{Z}_{\geq 0}^{n}$.

I am interested in the cardinality of the set ${\{\alpha \in \mathbb{Z}_{\geq 0}^{n} : |\alpha| \leq N}\}$, where $|\alpha| = a_1 + a_2 + ... + a_n$.

Does anyone know how to prove this? I assume there is some sort of combinatorial argument but I'm stuck? Any hints would be appreciated.

the man
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3 Answers3

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Let $A(n, N)$ be your number. Define $A_r(n, N)$ to be the cardinality of the set $$ \{\alpha\in \Bbb Z_{\geq 0}^n\mid |\alpha|\leq N, a_n = r\} $$ Clearly, we have $$ A(n, N) = \sum_{r = 0}^N A_r(n, N) $$ Also, note that by simply removing the last element of $\alpha$, we have $$ A_r(n, r) = A(n-1, N-r) $$ which is to say $$ A(n, N) = \sum_{r = 0}^N A(n-1, N-r) $$ Comparing the sum for $A(n, N)$ and the sum for $A(n, N-1)$, we get $$ A(n, N) = A(n-1, N) + A(n, N-1) $$ and we see that these are just relabelled binomial cefficients: $$ A(n, N) = \binom{n+N}{N} $$

Arthur
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  • By the stars and bars method, the number of non-negative $n$-tuples which sum to $N$ is $\binom{N+n -1}{N}$? This is the number you've worked out for the cardinality of my set? But the cardinality of my set must be bigger than this right? – the man Mar 28 '19 at 15:26
  • It must? Have you checked? For $n = 1$ or $n = 2$ it should be a simple thing to count by hand and compare. – Arthur Mar 28 '19 at 15:28
  • My set includes the case where the non-negative $n$-typles sum to $N$. But it also includes the non-negative $n$-tuples which sum to $0,1,...,N-1$. – the man Mar 28 '19 at 15:29
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    @theman You're right. I'm off by 1 somewhere. Let me just recheck my calculations. – Arthur Mar 28 '19 at 15:34
  • @theman There we go. I had the wrong base case ($n = 0$ is valid, and $A(0,N) = 1$), and also I hadn't looked thoroughly enough at my sums. Now it should be correct. – Arthur Mar 28 '19 at 15:47
  • Ok, thanks a lot for your help! – the man Mar 28 '19 at 16:01
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Hint: for $\alpha = N$ you are looking for weak compositions of $N$ into $n$ parts. You can solve that with the usual stars and bars argument. Now just sum the number of compositions of $k$ from $0$ to $N$

Ross Millikan
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Think of this as putting $N$ items into $n+1$ bins, where $a_1,\dots,a_n$ are the contents of the first $n$ bins. Using stars and bars, this gives $\binom{N+n}{n}$.

robjohn
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