How many positive integer solutions are there to the equation $(a + b + c + d) < N$? 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?
 A: 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.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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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:


