TITLE: How many lists of 100 numbers (1 to 10 only) add to 700?
QUESTION [34 upvotes]: Each number is from one to ten inclusive only.  There are $100$ numbers in the ordered list. The total must be $700$.
How many such lists?
Note: if, as it happens, this is one of those math problems where only an approximation is known, that would be great.  (My guess is it's not "that" big, around $10^{20}$ maybe?)
Thank you!!

So, to be clear say you have a dice with sides labelled $1$-$10$. You roll it $100$ times, once every $10$ seconds in order. The result is, if you will, a specific array of $100$ numbers (each being $1$-$10$), each position labelled $1$ to $100$. The array must then add to $700$; how many such arrays??

Just to be absolutely clear, I believe the total of "all" such lists (so, there is no requirement to add to $700$; it can add to anything), is indeed, simply $1$ googol, ie, $10^{100}$.

REPLY [6 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With $\ds{0 < a < 1}$:

\begin{align}
&\color{#66f}{\large\sum_{k_{1}=1}^{10}\ldots\sum_{k_{100}=1}^{10}
\delta_{k_{1}\ +\ \cdots\ +\ x_{100},700}}
=\sum_{k_{1}=1}^{10}\ldots\sum_{k_{100}=1}^{10}\oint_{\verts{z}\ =\ a}
{1 \over z^{-k_{1}\ -\ \cdots\ -\ k_{100}\ +\ 701}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ a}
{1 \over z^{701}}\,\pars{\sum_{k =1}^{10}z^{k}}^{100}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ a}{1 \over z^{701}}\,\pars{z\,{z^{10} - 1 \over z - 1}}^{100}
\,\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ a}{1 \over z^{601}}\,
{\pars{1 - z^{10}}^{100} \over \pars{1 - z}^{100}}\,\,{\dd z \over 2\pi\ic}
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\pars{1}
\\[3mm]&=\sum_{n = 0}^{100}\sum_{k = 0}^{\infty}\pars{-1}^{n + k}{100 \choose n}
{-100 \choose k}\oint_{\verts{z}\ =\ a}{1 \over z^{601 - 10n - k}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{n = 0}^{100}\sum_{k = 0}^{\infty}\pars{-1}^{n + k}{100 \choose n}
\pars{-1}^{k}{k + 99 \choose 99}\delta_{10n + k,600}
=\sum_{n = 0}^{100}\pars{-1}^{n}{100 \choose n}{699 - 10n \choose 99}
\end{align}

Contributions to the sum are limited by $\quad\ds{0 \leq n \leq 100}\quad$
and $\quad\ds{99 \leq 699 - 10n}\quad$ which yield $\quad\ds{0 \leq n \leq 60}$:
$$
\color{#66f}{\large\sum_{k_{1}=1}^{10}\ldots\sum_{k_{100}=1}^{10}
\delta_{k_{1}\ +\ \cdots\ +\ x_{100},700}
=\sum_{n = 0}^{60}\pars{-1}^{n}{100 \choose n}{699 - 10n \choose 99}}
$$
which is $\ds{\pars{~\mbox{see expression}\ \pars{1}~}}$ equal to $\ds{\bracks{z^{600}}\pars{1 - z^{10} \over 1 - z}^{100}}$.

It leads to the value
  \begin{align}&
{\tt 1.211306391075837746814517416259229640857139892912601984348493171327624}
\\
&{\tt 03014516376282321342995 \times 10^{92}}
\end{align}

It is found with W & A.