Number of $n$-tuples of positive integers summing to $k$

Asked May 10 '24 at 22:05

Active May 10 '24 at 22:05

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I know that:

$$\left|\left\{(a_1,a_2,...,a_n) \in \mathbb Z_{\ge0}^n : \sum_ia_i = k\right\}\right| = \binom{n + k - 1}k$$

In other words, the number of ordered $n$-tuples of non-negative integers summing to $k$ is given by the binomial coefficient on the RHS.

I tried to figure out a way to find the formula in the case of tuples of $\mathbb Z_{>0}^n$, that is, positive integers instead of non-negative integers, but I couldn't figure out such formula. The case $n=2$ is trivial, since it's enough to subtract $2$ from the result (namely, the pairs $(n,0)$ and $(0,n)$), but in the general case it gets very hard and I believe the formula doesn't look nice (although I'd love to get proven wrong on this point).

So, can you help me finding such formula? Thanks in advance.

asked May 10 '24 at 22:05

Elvis

1,543

2

Hint: If $(a_1,\dots,a_n)$ are positive, then $(a_1-1,a_2-1,\dots,a_n-1)$ are non-negative. – Thomas Andrews May 10 '24 at 22:32
How can it be $\binom{n-1}{k-n}$ if $k-n>n-1$ in some cases? – Elvis May 10 '24 at 22:47
1

Sorry, I thought I deleted the first one. Deleted whole comment. It is $\binom{k-1}{k-n}=\binom{k-1}{n-1}.$ If $n>k, $ there are no such tuples. – Thomas Andrews May 10 '24 at 22:56
Makes sense, thanks! – Elvis May 10 '24 at 23:02
See also this answer. In that answer, ignore all of the Inclusion-Exclusion discussion, review the Stars and Bars links, and then specifically see Addendum-2, at the end of the linked answer. – user2661923 May 11 '24 at 01:29

Number of $n$-tuples of positive integers summing to $k$

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