Distributing money to 4 people In how many ways can I collect a total of 20 dollars from 4 people, if each person can give any number of dollars from 0 to 10?
I used stars and bars for this question, but I'm not sure if my answer is correct. Since there are 20 dollars, there would be 19 bars and 19+4 = 23 total items. So, 23 Choose 4 = 8855, which looks really big so I'm not sure if this is correct.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$\ds{\bbox[5px,#ffd]{}}$

The answer is given by
\begin{align}
&\bbox[5px,#ffd]{\sum_{p_{1} = 0}^{10}
\sum_{p_{2} = 0}^{10}\sum_{p_{3} = 0}^{10}
\sum_{p_{4} = 0}^{10}\bracks{z^{20}}z^{p_{1}\ +\ p_{2}\ +\ p_{3}\ +\ p_{4}}} =
\bracks{z^{20}}\pars{\sum_{p = 0}^{10}z^{p}}^{4}
\\[5mm] = &\
\bracks{z^{20}}\pars{z^{11} - 1 \over z - 1}^{4} =
\bracks{z^{20}}\pars{1 - z^{11}}^{4}\pars{1 - z}^{-4}
\\[5mm] = &\
\bracks{z^{20}}\pars{1 - 4z^{11}}\pars{1 - z}^{-4} \\[5mm] = &\
\bracks{z^{20}}\pars{1 - z}^{-4} -
4\bracks{z^{9}}\pars{1 - z}^{-4}
\\[5mm] = &\
{-4 \choose 20}\pars{-1}^{20} -
4{-4 \choose 9}\pars{-1}^{9}
\\[5mm] = &\
\underbrace{{23 \choose 20}\pars{-1}^{20}}_{\ds{1771}}\ +\
4\ \underbrace{{12 \choose 9}\pars{-1}^{9}}_{\ds{-220}} =
\bbx{891} \\ &
\end{align}
A: It's not correct for a couple reasons, but it's a good place to start.  Generally the stars are used to count how many items there are to be distributed while the bars are separators.  So there are 20 stars for the 20 dollars and 3 bars to separate 4 people.  You would have $\binom{23}3$.
The problem is the count for stars and bars include the cases where one person gives 11 dollars or more.  Let's say we want to exclude the number of cases where person 1 gives 11 dollars or more.  That leaves 9 remaining dollars coming from 4 people.  To count the number of ways to collect 9 dollars from 4 people would be $\binom{12}3$ (9 stars, 3 bars).  The number will be the same if either of the other 3 people overspends and it is impossible for more than one person to overspend.  The result would then be $\binom{23}3-4\binom{12}3$.
