Solution from Intuition on Bose-Einstein Counting I am trying to show that to calculate the number of distinguishable arrangements of $N$ particles in $n$ boxes is $C(n-1+N, N)$ when using the Bose-Einstein particles. Could someone give me a solution?
The following is my attempt. 
Bose-Einstein
The particles are all identical, and I can put multiple particles in a given box. Because the particles are identical, each arrangement has an equal probability.
If there are $n$ boxes and $N$ particles then I have at least $C(n, N)$. Although I am missing all of the arrangements where there are multiple particles in a box. 
That means I am missing $C(n, N-1), C(n, N-2), ... , C(n, N-x)$ for all $x \in \mathbb N < N$. Each $C(n, N-x)$, represents that $x$ particles are grouped up with the other particles.
Now I have:
\begin{align}
\sum_{x=0}^N C(n, N-x) \tag{1} \label{1}
\end{align}
equals the number of possible arrangements.
Conclusion
I do not know to go from ($\ref{1}$) to $C(n-1+N, N)$, could someone help?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\sum_{\mathsf{box}_{1} = 0}^{\infty}\ldots\sum_{\mathsf{box}_{n} = 0}^{\infty}
\bracks{z^{N}}z^{\mathsf{box}_{1} + \cdots + \mathsf{box}_{n}} =
\bracks{z^{N}}\pars{\sum_{\mathsf{box} = 0}^{\infty}z^{\mathsf{box}}}^{n} =
\bracks{z^{N}}\pars{1 \over 1 - z}^{n}
\\[5mm] = &\
\bracks{z^{N}}\pars{1 - z}^{-n} = {-n \choose N}\pars{-1}^{N} =
\braces{{-\bracks{-n} + N - 1 \choose N}\pars{-1}^{N}}\pars{-1}^{N}
\\[5mm] = &\ \bbx{N + n - 1 \choose N}
\end{align}

Stars and Bars works too !!!.

