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Note: I know there's a very similar question to this one, where only positive integers are counted, but it's not helpful enough for me to answer my question at the bottom.

My approach:

1. Stars and Bars problem, so we can do as follows:

$x_1 + ..... + x_n = k$

Which results in $ k+n-1 \choose n-1$

2. We can continue like this:

$x_1 + ..... + x_n = k-1 $
$ k+n-2 \choose n-1$

$x_1 + ..... + x_n = k-2 $
$ k+n-3 \choose n-1$

All the way down to

$x_1 + ..... + x_n = k-k $
$ k+n-k-1 \choose n-1$

Which results in the following sum: $\sum_{i=0}^{k} {k+n-i-1 \choose n-1}$

While searching online for the correct answer, the answer key gave me this: $n+k \choose n$. Unfortunately, it doesn't give much explanation as to why this is true.

Okay, so my question is as follows: Is my approach/answer correct? If not, why does it not work? Because this sounds extremely intuitive to me.
If it is correct, how can I show that my answer is equal to that of the answer key? i.e. $\sum_{i=0}^{k} {k+n-i-1 \choose n-1} = {n+k \choose n}$

MrPuffer
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  • "I know there's a very similar question to this one, where only positive integers are counted, but it's not helpful enough for me to answer my question at the bottom": 1) "The bottom" should not be the title. Please edit your post: copy the title at the beginning of the body. 2) On the contrary, it is extremely helpful. Let $y_i:=x_i+1$ and consider the "vectors" $(y_1,\dots,y_n)$ such that $\sum y_i\le k+n$. – Anne Bauval Jun 24 '24 at 15:40
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    Does this answer your question? Prove $\sum\limits_{i=0}^n\binom{i+k-1}{k-1}=\binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity) Your approach is correct, and that identity shows that $\sum_{i=0}^{k} {k+n-i-1 \choose n-1} = {n+k \choose n}$. – Anne Bauval Jun 24 '24 at 15:56
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    The alternative approach, which is much easier, is to enumerate the number of solutions to $$x_1 + x_n + c = k ~: ~c \in \Bbb{Z_{\geq 0}}.$$ – user2661923 Jun 24 '24 at 15:58

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