Here's my attempt:
My thinking is that this is the same as finding all the non-negative $a, b, c, d$ such that $a + b + c + d = M$ where $M \in \{0, 1, ..., N - 4\}$. Which further reduces to a stars and bars problem, thus we get:
$\sum\limits_{k=0}^{N-4} {{k+4-1} \choose {k}}$.
But this seems to be incorrect. Can someone tell me why?
On
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Obviously, the desired result vanishes out when $\ds{N \leq 4}$. So, we consider the case $\ds{N \geq 5}$.
The number of combination with $\ds{a + b + c + d = s}$ $\ds{\pars{~s \geq 4~}}$ is given by: \begin{align}&\color{#66f}{\large% \sum_{a\ =\ 1}^{\infty}\sum_{b\ =\ 1}^{\infty} \sum_{c\ =\ 1}^{\infty}\sum_{d\ =\ 1}^{\infty}\delta_{a + b + c + d,s}} =\sum_{a\ =\ 1}^{\infty}\sum_{b\ =\ 1}^{\infty} \sum_{c\ =\ 1}^{\infty}\sum_{d\ =\ 1}^{\infty} \oint_{\verts{z}\ =\ 1^{-}}{1 \over z^{-a - b - c - d + s + 1}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1^{-}} {1 \over z^{s + 1}}\pars{\sum_{a\ =\ 1}^{\infty}z^{a}}^{4}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ 1^{-}}{1 \over z^{s + 1}}\pars{z \over 1 - z}^{4} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1^{-}} {1 \over z^{s - 3}\pars{1 - z}^{4}}\,{\dd z \over 2\pi\ic} =\sum_{k\ =\ 0}^{\infty}{-4 \choose k}\pars{-1}^{k}\oint_{\verts{z}\ =\ 1^{-}} {1 \over z^{s - 3 - k}}\,{\dd z \over 2\pi\ic} \\[5mm]&=\sum_{k\ =\ 0}^{\infty}{-4 \choose k}\pars{-1}^{k}\delta_{s - 3 - k,1} ={-4 \choose s - 4}\pars{-1}^{s}={s - 1 \choose s - 4} \\[5mm]&=\color{#66f}{\large{\pars{s - 1}\pars{s - 2}\pars{s - 3} \over 6}} \end{align}
The number of combination with $\ds{a + b + c + d < N}$ $\ds{\pars{~N \geq 5~}}$ is given by:
\begin{align}&\color{#66f}{\large% \sum_{s\ =\ 4}^{N - 1}{\pars{s - 1}\pars{s - 2}\pars{s - 3} \over 6}} =\color{#66f}{\large{N^{4} - 10N^{3} + 35N^{2} - 50N + 24 \over 24}} \end{align} with $\ds{N \geq 5}$ and zero otherwise $\ds{\pars{~N < 5~}}$.
A few values are given by:
I have $N$ doughnuts and want to distribute $\lt N$ of them among $4$ people, A, B, C, and D, with everyone getting at least $1$ doughnut. So I will keep the rest to eat myself. The number of ways to do this is the number of ways to distribute exactly $N$ doughnuts among $5$ people, A, B, C, D, and I, with everyone getting at least $1$.
So I line up the $N$ doughnuts, like this, with a little space between them. $$ O\quad O \quad O \quad O\quad O \quad O \quad O\quad O \quad O \quad O\quad O \quad O \quad O\quad O $$ There are $N-1$ interdoughnut gaps, and I choose $4$ of them to put separators into. A will get the doughnuts from the left end to the first separator, B will get the doughnuts from the first separator to the second, and so on. I will get the ones from the last separator to the right end. Choosing where the separators go can be done in $\binom{N-1}{4}$ ways.