"Story" proof of $\sum_{i=j}^n {i \choose j} = {n+1 \choose j+1}$ I am reading a book "Discrete Mathematics for Computer Scientists". One of the exercises asks for a "story" proof of this: $\sum_{i=j}^n {i \choose j} = {n+1 \choose j+1}$.

My question is that:


What is a story proff?
What is the story proof of this identity?


Screenshot of the book:
If you read ${n+1} \choose{j+1}$ as: You have $n+1$ numbered boxes $\{1,...,n+1\}$ and you want to place $j+1$ indistinguishable balls in these boxes (at most one in a box), how many ways is that possible . 
Then we look at different cases: To do this look at $x$ defined as the highest number for which the box is occupied. Note $x&gt;j$ since all balls must be placed. This means that the other $j$ balls are distributed in the boxes $\{1,...,x-1\}$. This can be done in ${x-1} \choose {j}$ ways. Now we can sum $x$ over ${j+1,...,n+1}$. Therefor we get $\sum _{i=j}^n {i\choose j}={n+1 \choose j+1}$.
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\begin{align}&amp;\color{#66f}{\large\sum_{k\ =\ j}^{n}{k \choose j}}
=\sum_{k\ =\ j}^{n}\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{k} \over z^{j + 1}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}
{1 \over z^{j + 1}}\sum_{k\ =\ j}^{n}\pars{1 + z}^{k}\,{\dd z \over 2\pi\ic}
\\[5mm]&amp;=\oint_{\verts{z}\ =\ 1}{1 \over z^{j + 1}}\,\pars{1 + z}^{j}\,
{\pars{1 + z}^{n - j + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic}
\\[5mm]&amp;=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n + 1}\over z^{j + 2}}
\,{\dd z \over 2\pi\ic}
-\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{j}\over z^{j + 2}}
\,{\dd z \over 2\pi\ic}
={n + 1 \choose j + 1} -\
\underbrace{j \choose j + 1}_{\ds{=\ \color{#c00000}{0}}}\ = \
\color{#66f}{\large{n + 1 \choose j + 1}}
\end{align}